\documentclass[11pt]{article} \usepackage[a4paper,left=27mm,right=27mm,top=25mm,bottom=25mm]{geometry} \usepackage[T1]{fontenc} \usepackage{lmodern,amsmath,amssymb,amsthm,mathtools,mathrsfs} \usepackage{booktabs,tabularx,array,enumitem,microtype,xcolor,hyperref,needspace,longtable} \setlist[enumerate]{itemsep=3pt,topsep=5pt,parsep=0pt} \definecolor{inkblue}{RGB}{25,55,85} \hypersetup{hidelinks, pdftitle={Vertex Operator Algebra Realizations of Twisted Haagerup-Izumi Doubles}, pdfauthor={David E. Evans and Andrew Riesen},pdfsubject={Regular nonunitary VOA realizations of twisted Haagerup-Izumi centers}} \pagestyle{plain} \setlength{\parindent}{1.2em}\setlength{\parskip}{0pt} \setlength{\emergencystretch}{2em} \setcounter{tocdepth}{1} \setcounter{secnumdepth}{2} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newcommand{\C}{\mathbb C}\newcommand{\Q}{\mathbb Q}\newcommand{\Z}{\mathbb Z} \newcommand{\one}{\mathbf1}\newcommand{\D}{\mathcal D}\newcommand{\Dcat}{\mathcal D} \newcommand{\Ncat}{\mathcal N}\newcommand{\Pcat}{\mathcal P}\newcommand{\Fcat}{\mathcal F} \newcommand{\Scat}{\mathcal S}\newcommand{\U}{\mathcal U}\newcommand{\bo}{\boxtimes} \newcommand{\FP}{\operatorname{FPdim}}\newcommand{\Irr}{\operatorname{Irr}} \newcommand{\rank}{\operatorname{rank}}\newcommand{\End}{\operatorname{End}} \newcommand{\Hom}{\operatorname{Hom}}\newcommand{\id}{\operatorname{id}} \newcommand{\Vir}{\operatorname{Vir}}\newcommand{\Rep}{\operatorname{Rep}} \newcommand{\Ext}{\operatorname{Ext}}\newcommand{\wt}{\operatorname{wt}} \newcommand{\Spec}{\operatorname{Spec}}\newcommand{\soc}{\operatorname{soc}}\newcommand{\ch}{\operatorname{ch}} \newcommand{\N}{\mathcal N}\newcommand{\F}{\mathcal F} \newcommand{\zetaM}[1]{\zeta_{#1}} \numberwithin{equation}{section} \newcommand{\R}{\mathbb R} \begin{document} \title{Vertex Operator Algebra Realizations of\\Twisted Haagerup--Izumi Doubles} \author{David E. Evans \and Andrew Riesen} \date{22 October 2026} \maketitle \begin{abstract} We construct vertex operator algebras realizing an infinite family of twisted, nonunitary Haagerup--Izumi doubles. For every odd integer $N\geq3$, we construct a spherical fusion category $\Fcat_N$ with the Haagerup--Izumi fusion rules for $\Z/N\Z$ and pointed associator of exact order $N$, and realize its Drinfeld center as the ribbon category of modules of an explicit vertex operator algebra $\mathscr{V}_N$. The construction begins with a uniform family of strongly rational Virasoro extensions, whose irreducible modules and theta-function characters we determine. A simple-current extension involving a conformally shifted $E_7$ lattice vertex operator algebra removes the additional semion factor. The resulting $\mathscr{V}_N$ is simple, regular, self-contragredient, and $C_2$-cofinite, with central charge $-6N^2-10$, effective central charge eight, and $N^2+3$ irreducible modules with explicit characters. Its weight-zero subspace has dimension $56$, so it is not of CFT type; we establish modularity and the ribbon equivalence directly. The uniform rationality argument combines singular-vector recurrences, a $2$-adic finiteness obstruction, and hypergeometric control of the Zhu algebra. \end{abstract} \section{Introduction} The realization of Haagerup--Izumi doubles by vertex operator algebras is part of the broader problem of realizing modular tensor categories by rational conformal field theories. The first author and Gannon~\cite{B:EG11} introduced families of twisted Haagerup--Izumi modular data and studied their possible VOA realizations. In this paper, we construct vertex operator algebras realizing an infinite family of these twisted nonunitary doubles, with a ribbon equivalence identifying the representation categories. Recent work has also advanced the realization problem for untwisted doubles. Gannon, Schopieray and Yadav~\cite{B:GSY} construct Haagerup--Izumi fusion categories for every odd cyclic group and determine the modular data of their centers. Lattice constructions for the untwisted Haagerup double at central charge eight appear in work of the second author~\cite{I:Riesen3} and T.-C. Huang~\cite{I:Huang3}. Huang~\cite{I:Huang5} also constructs a unitary strongly rational VOA realizing the positive order-five Haagerup--Izumi double inside $V_{A_4\oplus A_4}$. Our starting point is the diagonal Virasoro construction of the first author~\cite{A:Evans}. It was motivated by the S-fold theories of Gang and Kim~\cite{I:GK} and the boundary characters proposed by Gang, Kim and Lee~\cite{I:GKL}; its theta-function characters are also related to Flohr's earlier work on two-generator $\mathcal W$-algebras~\cite{I:Flohr}. For odd $n\geq3$, the underlying Virasoro module is \begin{equation}\label{A:construction} V=V_{n+2}=\bigoplus_{j\geq0}L(1-6n,nj(j+1)), \end{equation} where $L(c,h)$ denotes the simple Virasoro module of central charge $c$ and lowest weight $h$. The multiplication is the canonical diagonal multiplication of \cite[Theorem~3.2]{A:Evans}, recalled in Section~\ref{sec:diagonal}. The first author established strong rationality and identified the twisted center factor at $n=9$, and formulated the uniform extension of these results in \cite[Conjecture~11.1]{A:Evans}. We establish the structure of this family for every odd $n\geq3$. In particular, we prove strong rationality, classify the irreducible modules and determine their characters. If $J$ denotes the order-two simple current, write \[ \Scat_n=\langle J\rangle,\qquad \D_n=\operatorname{Cent}_{\Rep(V_{n+2})}(J),\qquad m=n+4. \] There is a ribbon factorization \begin{equation}\label{E:factorintro} \Rep(V_{n+2})\simeq\Scat_n\bo\D_n, \end{equation} where $\Scat_n$ is a semion category with $d(J)=-1$. A lattice extension of $V_{n+2}$ determines a connected \'etale algebra in $\D_n$, whose module category has near-group fusion rules. For $n=N^2$, a further condensation constructs a spherical fusion category $\Fcat_N$ with Haagerup--Izumi fusion rules and gives a ribbon equivalence $\D_{N^2}\simeq Z(\Fcat_N)$. More generally, $\D_n$ is a Drinfeld center exactly when $n$ is a square. The remaining step is to realize $\D_{N^2}$ itself by a VOA. We use a simple-current extension to cancel the semion factor in \eqref{E:factorintro}, following the reconstruction method of Gannon and the second author~\cite[Theorem~2.3]{E:GR}. The spherical structure is essential: the auxiliary semion must have the braiding and twist compatible with the negative dimension of $J$. A minuscule conformal shift of the $E_7$ lattice VOA provides the required inverse ribbon factor and integral extension weights. This gives the following main result. \begin{theorem}[Twisted-center realization]\label{E:main} Let $N\geq3$ be odd and put $n=N^2$. Let $K=E_7$, let $u\in K^*\setminus K$ be a minuscule weight with $u^2=3/2$, and set \[ \mathcal W=(V_K,\omega_K+u(-2)\one),\qquad M=(V_{K+u},\omega_K+u(-2)\one). \] In the second expression the conformal vector acts on the lattice module. Then \begin{equation}\label{E:def} \mathscr{V}_N=(V_{N^2+2}\otimes \mathcal W)\oplus(J\otimes M) \end{equation} admits a simple, regular, $C_2$-cofinite VOA extension structure, unique up to an isomorphism fixing $V_{N^2+2}\otimes \mathcal W$. It is self-contragredient and nonnegatively integer graded, and \begin{equation}\label{E:maininvariants} \begin{gathered} \Rep(\mathscr{V}_N)\simeq Z(\Fcat_N),\qquad |\Irr(\mathscr{V}_N)|=N^2+3,\\ c(\mathscr{V}_N)=-6N^2-10,\qquad \dim(\mathscr{V}_N)_0=56,\\ h_{\min}(\mathscr{V}_N)=-\frac{N^2+3}{4},\qquad c_{\mathrm{eff}}(\mathscr{V}_N)=8. \end{gathered} \end{equation} The equivalence is ribbon. The category $\Fcat_N$ is the category constructed in Section~\ref{A:sec:categorical}, with fusion rules \[ \alpha_a\alpha_b=\alpha_{a+b},\quad \rho\alpha_a=\alpha_{-a}\rho,\quad \rho^2=\one\oplus\bigoplus_{a\in\Z/N}\alpha_a\rho. \] Its noninvertible spherical dimension and pointed associator are \begin{equation}\label{E:HIinvariants} \begin{gathered} d(\rho)=\frac{N-\sqrt{N^2+4}}2,\\ \Omega_N(a,b,c)=\exp\!\left(\frac{2\pi i c}{N} \left\lfloor\frac{a+b}{N}\right\rfloor\right),\qquad 0\leq a,b,c1$ with $t+t^{-1}=n+2$. This number is irrational. At $c=13-6(t+t^{-1})=1-6n$, set \[ h_{r,s}=\frac{(rt-s)^2-(t-1)^2}{4t}. \] The category of $C_1$-cofinite Virasoro modules at this generic central charge is semisimple and rigid, with simple modules $L(c,h_{r,s})$ for $r,s>0$ \cite{A:CJORY}. Its two subcategories generated by the odd $r$ labels $L(c,h_{r,1})$ and the odd $s$ labels $L(c,h_{1,s})$ centralize each other. Indeed, \[ h_{r,s}-h_{r,1}-h_{1,s}=-\tfrac12(r-1)(s-1)\in2\Z \] for odd $r,s$, and $L(c,h_{r,1})\bo L(c,h_{1,s})\cong L(c,h_{r,s})$. The first subcategory is the generic integer-spin category $SO(3)_q$ with $q^2=e^{2\pi it}$; the second has parameter $q'^2=e^{2\pi i/t}=q^{-2}$. Thus the second is braid-reversed equivalent to the first, taking label $r$ to label $r$. The generic type-$A_1$ braiding and fusion rules give this equivalence, including the twists; on a diagonal label, $h_{r,1}+h_{1,r}\in\Z$. The canonical diagonal algebra in the product of a rigid category with its braid reverse has underlying object \[ \bigoplus_{r>0\;\mathrm{odd}}L(c,h_{r,1})^*\bo L(c,h_{1,r}) \cong\bigoplus_{j\geq0}L(c,nj(j+1)). \] Its multiplication is obtained by pairing dual bases of the fusion spaces; it is nonzero in every allowed channel. The gluing construction and its direct-limit extension theorem identify it with a VOA extension \cite{A:CKM,A:Ind}. It is $\Z_{\geq0}$-graded with finite-dimensional graded pieces and weight-zero space $\C\one$, so the VOA-extension correspondence applies. This VOA is simple. An ideal is a Virasoro submodule, hence a sum of diagonal simple summands. Multiplying a nonzero summand by its dual reaches the vacuum because the corresponding pairing is nonzero. Thus every nonzero ideal contains the unit. Let $W$ span the top of the $j=1$ summand, of weight $2n$. Fusion with this summand reaches $j+1$ from $j$, so $T,W$ generate the VOA. The singular $WW$ OPE has weights at most $4n-1$, below the next diagonal primary of weight $6n$. It therefore consists of descendants of the vacuum and $W$; closure of these generator OPEs proves strong generation. Finally $V_1=0$, so the invariant-form criterion for a simple CFT-type VOA \cite{A:Li} makes $V$ self-contragredient. \subsection{Notation} Write $M(c,h)$ for a Virasoro Verma module and $L(c,h)$ for its simple quotient. Vertex modes use $Y(a,z)=\sum_{m\in\Z}a_mz^{-m-1}$, $L_m=T_{m+1}$, $D=L_{-1}$, and $o(a)=a_{\wt(a)-1}$. In particular \[ C_2(V)=\operatorname{span}\{a_{-2}b:a,b\in V\},\qquad C_1(M)=\operatorname{span}\{a_{-1}m:\wt(a)>0\}. \] An ordinary module has diagonalizable $L_0$, finite-dimensional weight spaces, and weights bounded below. Generalized modules allow finite Jordan blocks. CFT type means $V_j=0$ for $j<0$ and $V_0=\C\one$. Regularity means that every weak module is a direct sum of ordinary irreducible modules. Strong rationality means simplicity, CFT type, self-contragredience, $C_2$-cofiniteness, and rationality. A ribbon equivalence is a braided tensor equivalence preserving the twist. For a spherical fusion category $\mathcal F$, $Z(\mathcal F)$ denotes its Drinfeld center with the induced ribbon structure. We reserve $\mathscr{V}_N$ for the final realizing VOA, write $V=V_{n+2}$ for the diagonal Virasoro extension, and use $V_L$ for the lattice VOA associated to an even positive-definite lattice $L$. The symbols $\FP$ and $d$ denote Frobenius--Perron and spherical dimensions; the latter may be negative. The rank of a fusion category is its number of simple isomorphism classes. The centralizer of $J$ consists of objects whose double braiding with $J$ is the identity. The symbol $\bo$ denotes a VOA fusion product or a Deligne product as its arguments dictate. All categories are over $\C$. The letter $W$ denotes the primary generator of $V$; the shifted $E_7$ lattice VOA used in Sections~\ref{E:auxiliary}--\ref{E:scope} is denoted by $\mathcal W$. For a commutative algebra $A$ in a braided category, an $A$-module is \emph{local} if its action is unchanged by double braiding with $A$. Write $\D_A$ for all $A$-modules and $\D_A^{\mathrm{loc}}$ for the local ones. An algebra is \emph{connected \'etale} if $\Hom(\one,A)=\C$ and its multiplication has an $A$-bimodule section. Passing to $A$-modules is the condensation used below. \subsection{Tensor convention}\label{sec:tensorconvention} For a commutative algebra $A$ in a braided category $\D$, $\D_A$ means right modules with left action on $M$ obtained from $c^{-1}_{M,A}:A\otimes M\to M\otimes A$. This is the convention of \cite[Section 3.3]{A:DMNO}. For an algebra $B$ in a center $Z(\Ncat)$, the quotient $\operatorname{Rep}_{\Ncat}(B)$ uses its specified central half-braiding $B\otimes M\to M\otimes B$ to define the left action. These two conventions agree under the condensation-in-stages identification used below. Using the opposite convention changes the displayed pointed cocycle to its inverse. \section{\texorpdfstring{Uniform $C_2$-cofiniteness}{Uniform C2-cofiniteness}}\label{A:app:c2} The generator $W$ has weight $2n$ and belongs to the Virasoro module $L(1-6n,h_{3,3})$. Its level-nine null vector gives a differential equation for the image of $Y(W,z)W$ at a point of the $C_2$-algebra. A nonzero point would produce a polynomial solution; a $2$-adic recurrence rules out such a solution. \begin{theorem}\label{A:c2:thm:main} For every positive odd $n$, the diagonal algebra $V_{n+2}$ is $C_2$-cofinite. If $n\geq5$, then its $C_2$-algebra $R_n$, generated by $t=[T]$ and $w=[W]$, satisfies \[ t^{2n+4}=t^{n+4}w=0,\qquad \dim R_n\leq3n+8. \] Its Zhu algebra is finite-dimensional and commutative, so every irreducible module has a one-dimensional lowest space. \end{theorem} The Zhu assertion follows from strong generation by $T,W$, centrality of $[T]$, and the surjection $R_n\twoheadrightarrow\operatorname{gr}A(V)$. The coefficients of the normalized level-nine singular vector and its differential recurrence are recorded in Appendix~\ref{app:null}. \subsection{The differential equation at a $C_2$-point} Strong generation gives a surjection $\C[t,w]\twoheadrightarrow R_n$. Since the weight-$4n$ space lies below the next Virasoro highest weight, reduction of $W_{(-1)}W$ gives \begin{equation}\label{A:c2:eq:quad} w^2=A t^{2n}+B t^n w \end{equation} for scalars $A,B$. At any point of $\Spec R_n$ with $t=0$, equation \eqref{A:c2:eq:quad} gives $w=0$. Thus a nonzero point must have $t\ne0$. By the grading action it can be normalized to $t=1$. Suppose such a point exists, and let $\ell:V_{n+2}\to R_n\to\C$ be its evaluation functional. Then \[ \ell L_{-2}=\ell,\qquad \ell L_{-1}=0,\qquad \ell L_{-r}=0\quad(r\geq3). \] Set $F(z)=\ell(Y(W,z)W)$. Every positive power of $z$ has coefficient in $C_2(V)$, and odd-weight classes vanish because $R_n$ is generated in even weights. Hence \begin{equation}\label{A:c2:eq:finite} F(z)\in\C[z^{-2}],\qquad F(z)=\nu z^{-4n}+\cdots+w^2,\quad \nu\ne0. \end{equation} The nonzero leading coefficient is the nonzero vacuum pairing of the $W$ summand. For a descendant of the inserted field, the Virasoro Ward identities and the displayed properties of $\ell$ give \begin{align*} D_1&=\partial_z,\\ D_r&=\delta_{r,2}+(-1)^{r-2} \big((r-1)h z^{-r}-z^{1-r}\partial_z\big),\qquad r\geq2. \end{align*} These formulas also hold when the inserted vector is already a descendant. They follow by normal ordering: only $L_{-2}$ survives on the left, while $L_0$ and $L_{-1}$ are the only nonzero nonnegative-side modes on the right highest-weight vector. Let $s_{3,3}=\sum_{\lambda\vdash9}c_\lambda(n)L_{-\lambda}W$ be the level-nine singular vector, normalized by $c_{(1^9)}=1$. Define $P_j(e,n)$ by \begin{equation}\label{A:c2:eq:defP} \sum_{\lambda\vdash9}c_\lambda(n) D_{\lambda_1}\cdots D_{\lambda_\ell}z^e =\sum_{j=0}^4P_j(e,n)z^{e-9+2j}. \end{equation} The coefficients are listed in Appendix~\ref{app:null}. Commuting each $L_m$, $1\leq m\leq9$, to the right gives $L_m s_{3,3}=0$ as a polynomial identity in $n$. Since this singular vector vanishes in $L(c,2n)$, \begin{equation}\label{A:c2:eq:L} \mathcal L_n F=0,\qquad \mathcal L_n=\sum_{j=0}^4z^{-9+2j}P_j(z\partial_z,n). \end{equation} In particular, \[ P_4(e,n)=256en(n+4). \] With $x=z^{-2}$ and $\Theta=x\partial_x$, multiplication of \eqref{A:c2:eq:L} by $z$ gives the polynomial-coefficient operator \begin{equation}\label{A:c2:eq:A} \mathcal A_n=\sum_{j=0}^4x^jR_j(\Theta,n),\qquad R_j(k,n)=P_{4-j}(-2k,n). \end{equation} Thus \eqref{A:c2:eq:finite} would be a nonzero polynomial in the kernel of $\mathcal A_n$. \subsection{The $2$-adic obstruction} Write $D=n(n+4)$. The diagonal coefficient is \[ R_0(k,n)=-512Dk. \] Since $D\ne0$, a nonzero formal power-series solution at $x=0$ must have nonzero constant term and is uniquely determined by that term. Normalize it as \[ C(x)=\sum_{a\geq0}C_a x^a,\qquad C_0=1. \] The coefficients are rational and satisfy \begin{equation}\label{A:c2:eq:rec} 512aD\, C_a=\sum_{j=1}^{\min(4,a)}R_j(a-j,n)C_{a-j}. \end{equation} \begin{lemma}[Integer divisibilities]\label{A:c2:lem:div} For odd $n$ and integral $k$, \[ 512\mid R_1(k,n),\quad256\mid R_2(k,n),\quad128\mid R_3(k,n). \] If $4\mid k$, then $512\mid R_2(k,n)$. Furthermore \begin{align*} R_4(k,n)={}&-32(k+n)(n-k)(2n-k)A(k,n)B(k,n)C(k,n),\\ A(k,n)={}&4k^2+4k-4n^2-16n+1,\\ B(k,n)={}&k^2-kn+4k-2n^2-11n+4,\\ C(k,n)={}&4k^2-12kn+4k+8n^2-10n+1. \end{align*} Each of $A,B,C$ is odd for odd $n$ and integral $k$. \end{lemma} \begin{proof} The recurrence polynomials obtained from \eqref{A:c2:eq:defP} have contents $512,256,128,64,32$ for $R_0,\ldots,R_4$, respectively. For odd $n$, their primitive parts give \[ \frac{R_1}{256}\equiv k(k+1),\qquad \frac{R_2}{128}\equiv k(k+1),\qquad \frac{R_3}{64}\equiv k^4+k\pmod2. \] All three expressions vanish at every integral $k$. Direct substitution $k=4v$ into $R_2$ gives the additional divisibility by $512$. The factorization of $R_4$ is an exact polynomial identity. The parity assertions follow immediately, including $B\equiv k^2-k-1\equiv1\pmod2$. These divisibilities can be verified by expansion after $n=2u+1$ and $k=2v+r$ (or $k=4v$): coefficient divisibility in $\Z[u,v]$ gives the assertions on every residue class. This is a finite polynomial identity check, independent of any specialization of $n$. \end{proof} Let $K=\Q_2(\pi)$, $\pi^2=2$, and let $\mathcal O=\Z_2[\pi]$ be its valuation ring. Its residue field is $\mathbb F_2$. Define \[ b_a=\pi^a a!C_a. \] Equation \eqref{A:c2:eq:rec} becomes \begin{equation}\label{A:c2:eq:scaled} b_a=\sum_{j=1}^{\min(4,a)}q_j(a,n)b_{a-j},\qquad q_j=\frac{\pi^j(a-1)\cdots(a-j+1)R_j(a-j,n)}{512D}. \end{equation} The empty product for $j=1$ is $1$. Since $n$ is odd, $D$ is a unit. \Needspace{8\baselineskip} \begin{lemma}\label{A:c2:lem:units} Every coefficient $q_j(a,n)$ in \eqref{A:c2:eq:scaled} lies in $\mathcal O$. Moreover: \begin{enumerate} \item $q_1,q_3\in\pi\mathcal O$ for every applicable $a$; \item $q_2\in\pi\mathcal O$ unless $4\mid a$; \item $q_4\equiv1\pmod\pi$ if $4\mid a$, and $q_4\equiv0\pmod\pi$ otherwise. \end{enumerate} \end{lemma} \begin{proof} The $q_1$ assertion follows from $512\mid R_1$. For $q_3$, use $128\mid R_3$ and the even factor in $(a-1)(a-2)$: the resulting valuation is at least that of $\pi$. For $q_2=(a-1)R_2(a-2,n)/(256D)$, integrality follows from $256\mid R_2$. If $a$ is odd, $a-1$ is even. If $a\equiv2\pmod4$, then $a-2\equiv0\pmod4$, giving the extra factor of $2$ in Lemma~\ref{A:c2:lem:div}. For $q_4$, the exact factorization gives the following lower bounds, with equality in the first row. Here $v_2(2)=1$: \[ \begin{array}{c|c|c|c} a\bmod4 & v_2(R_4(a-4,n)) & v_2((a-1)(a-2)(a-3)) & v_2(q_4)\\\hline 0&6&1&0\\ 2&\geq7&\geq2&\geq2\\ 1,3&\geq8&\geq3&\geq4 \end{array} \] For instance, in the first row $a-4$ is divisible by $4$, the factors $k+n$ and $n-k$ are odd, and $v_2(2n-k)=1$. A unit has residue $1$ in $\mathbb F_2$. Zero factors only increase the lower bounds. \end{proof} \begin{proposition}\label{A:c2:prop:no-poly} For every positive odd integer $n$, the only polynomial solution of $\mathcal A_nF=0$ over $\C$ is zero. \end{proposition} \begin{proof} By Lemma~\ref{A:c2:lem:units} and induction, all $b_a$ lie in $\mathcal O$. The same induction modulo $\pi$ gives \begin{equation}\label{A:c2:eq:key} \displaystyle b_a\equiv \begin{cases}1&4\mid a,\\0&4\nmid a\end{cases}\pmod\pi. \end{equation} Indeed, if $a\not\equiv0\pmod4$, all applicable coefficients vanish modulo $\pi$. If $4\mid a$, the only additional possible term is $q_2b_{a-2}$, whose second factor is already zero modulo $\pi$; the fourth-step term propagates $b_{a-4}\equiv1$. Thus $C_{4m}\ne0$ for every $m\geq0$, so the normalized formal solution is not a polynomial. Any nonzero polynomial solution must have nonzero constant term, because $R_0(d,n)\ne0$ for every positive integer $d$; after normalization it would equal $C(x)$ coefficientwise. This proves the claim, also for complex coefficients. \end{proof} \subsection{Consequences for the $C_2$-algebra} Proposition~\ref{A:c2:prop:no-poly} contradicts the polynomial forced by any nonzero point in the preceding Ward-equation argument. Thus $\Spec R_n$ has only its origin. Since $R_n$ is a finitely generated commutative $\C$-algebra, it is zero-dimensional and finite-dimensional. This proves $C_2$-cofiniteness for every positive odd $n$. The same operator gives a linear bound on $\dim R_n$. \begin{proposition}\label{A:c2:prop:bound} For odd $n>4$, one has $t^{2n+4}=t^{n+4}w=0$ in $R_n$. \end{proposition} \begin{proof} On each of the vacuum and $W$ Virasoro channels, evaluate PBW monomials by sending pure $L_{-2}$ powers of the highest-weight vector to $1$ and every other negative-mode monomial to $0$. This is well defined: the negative-mode character kills $L_{-1}\mathbf1$, and kills the odd-level singular vector $s_{3,3}$ and all its descendants. The associated channel series are \[ F_0(z)=z^{-4n}\sum_{a\geq0}\alpha_a z^{2a},\qquad F_1(z)=z^{-2n}\sum_{a\geq0}\beta_a z^{2a},\qquad \alpha_0=\beta_0=1. \] Both satisfy $\mathcal L_nF_i=0$. Their distinct leading powers make them linearly independent. The $4\times2$ matrix \[ M_n=\begin{pmatrix} \alpha_{2n+1}&\beta_{n+1}\\ \alpha_{2n+2}&\beta_{n+2}\\ \alpha_{2n+3}&\beta_{n+3}\\ \alpha_{2n+4}&\beta_{n+4} \end{pmatrix} \] has rank two. Otherwise a nonzero combination $G$ of $F_0,F_1$ has zero coefficients of $z^2,z^4,z^6,z^8$. Its truncation $G_{\leq0}$ is a nonzero polynomial in $z^{-2}$, while $G-G_{\leq0}=O(z^{10})$. Since $\mathcal L_n$ lowers powers by at most nine, $\mathcal L_n(G-G_{\leq0})$ has only positive powers. Conversely $\mathcal L_nG_{\leq0}$ has only negative powers. The equation $\mathcal L_nG=0$ forces both to vanish, contradicting Proposition~\ref{A:c2:prop:no-poly}. For $j=1,\ldots,4$, the negative product $W_{(-2j-1)}W$ belongs to $C_2(V)$ and has weight $4n+2j<6n$. Its reduction therefore gives \[ \nu\alpha_{2n+j}t^{2n+j} +\mu\beta_{n+j}t^{n+j}w=0, \qquad \nu\mu\ne0, \] where $\nu,\mu$ are the two channel normalizations. Multiply the $j$th relation by $t^{4-j}$. Rank two now gives both claimed nilpotence relations. \end{proof} Together with \eqref{A:c2:eq:quad}, these relations span $R_n$ by \[ 1,t,\ldots,t^{2n+3},\qquad w,tw,\ldots,t^{n+3}w. \] There are $3n+8$ listed monomials, completing Theorem~\ref{A:c2:thm:main}. The rationality argument in Section~\ref{A:app:rat} addresses the further semisimplicity assertion. \Needspace{10\baselineskip} \section{The lattice realization}\label{A:app:lat} We construct the conformal lattice extension and the simple current used in the rationality argument. A charge filtration and the ordinary Virasoro block decomposition determine their irreducible summands and characters. The local-module calculation then identifies the pointed category that enters the categorical reconstruction. Write \[ c=1-6n,\qquad V=V_{n+2}=\bigoplus_{j\geq0}L(c,nj(j+1)),\qquad n\geq1. \] Let $L=\Z\beta$, $(\beta,\beta)=2n$, $H=\beta_{-1}\one$, $\varepsilon=e^\beta$, and $\lambda\ne0$. On the ordinary lattice vertex algebra put \begin{equation}\label{A:lat:stress} T_\lambda=\frac{1}{4n}H_{-1}^{2}\one+\frac12H_{-2}\one+\lambda D^2\varepsilon, \qquad \U=(V_L,T_\lambda). \end{equation} The symbols $\mathcal M_a$ denote the $2n$ lattice modules $V_{L+a\beta/(2n)}$. \begin{theorem}\label{A:lat:main} For every positive integer $n$, the following hold. \begin{enumerate}[label=(\roman*),leftmargin=7mm] \item There is a conformal embedding $V\hookrightarrow\U$. Its image is exactly the sum of the $C_1$-cofinite simple Virasoro submodules of $\U$. \item As $V$-modules, $\U=V\oplus X$ and $\mathcal M_n=J\oplus Y$, where $X,J,Y$ are simple and \[ J\big|_{\Vir}=\bigoplus_{j\geq1}L\left(c,n\left(j^2-\tfrac14\right)\right). \] The lowest weights of $X,J,Y$ are $0,3n/4,-n/4$, respectively. \item For $a\not\equiv0,n\pmod{2n}$, $\mathcal M_a$ is simple upon restriction to $V$, and $\mathcal M_a\cong \mathcal M_{2n-a}$ as $V$-modules. The restrictions with $1\leq a0$. The equality $k\beta=\alpha_{r,s}$ forces $r=s=1-2k$. Hence the Verma-to-Fock map is invertible for $k=j+1>0$. Fock contragredience sends $k$ to $1-k$, proving the other orientation in \eqref{A:lat:Fock}. In particular, $K_j$ is an irreducible Verma module and is not $C_1$-cofinite. Distinct $j$ give disjoint blocks. The only potential cross-equality would imply \[ n(2j+1)^2=(n+4)(2k+1)^2, \] which would make $n(n+4)=(n+2)^2-4$ a square. It is not a square for $n\geq1$. \begin{lemma}[Block projectors on weight-finite unions]\label{A:blocks:union} Suppose an ordinary lower-bounded Virasoro module $M$ has finite-dimensional weight spaces, and each of its finitely generated submodules has the functorial exact block decomposition in category $\mathcal O$. The block projectors extend compatibly to $M$, preserve its submodules, and remain exact on short exact sequences of such modules. \end{lemma} \begin{proof} A finitely generated submodule is in category $\mathcal O$: the positive modes lower weight, so their orbit on finitely many generators lies in a finite-dimensional sum of weight spaces. If $M_1\subset M_2$ are two such submodules, functoriality makes their block projections agree on $M_1$. Taking their directed union therefore defines a projector on $M$. Every vector has only finitely many block components because it belongs to one finitely generated submodule. A submodule is preserved by applying the same construction to its finitely generated submodules. For exactness, lift any finitely many vectors from a quotient to the middle module and generate a finitely generated Virasoro submodule there. Exactness of its block decomposition gives the required lifts of block components. This proves exactness of the extended projections. In particular, one may apply a block projector to the charge filtration and its quotients before taking the weightwise finite associated graded. This argument is restricted to ordinary modules; it is not a statement about generalized $L_0$-blocks. \end{proof} \begin{lemma}\label{A:lat:blocklemma} Under the deformed Virasoro action, \[ \U=\bigoplus_{j\geq0}B_j,\qquad 0\longrightarrow M_j\longrightarrow B_j\longrightarrow M_j^\vee\longrightarrow0. \] Each $B_j$ has precisely one simple $C_1$-cofinite submodule, isomorphic to $L_j$. Its highest vector $W_j$ has lowest charge $-j$, with nonzero coefficient of $e^{-j\beta}$. \end{lemma} \begin{proof} The filtration $F^r\U=\bigoplus_{k\geq r}\mathsf F_k$ is Virasoro-stable: the perturbation raises charge by one. It is finite and exhaustive on every conformal-weight space. Its graded quotients have the undeformed Fock action. Apply Lemma~\ref{A:blocks:union} to the charge filtration. In the $j$th block, only the two quotients \eqref{A:lat:Fock} remain. Weightwise finiteness of the filtration therefore gives the displayed exact sequence. The bottom space of $B_j$ is two-dimensional. Apply the Verma singular operator at level $(2j+1)^2$ to this space. It is nonzero on the bottom of the embedded $M_j$. Its image in $M_j^\vee$ is zero. Its image in $M_j$ consists of singular vectors, a one-dimensional space. Thus the map has rank exactly one. Its one-dimensional kernel generates $L_j$; no other $C_1$-cofinite simple can occur in this block. A kernel vector cannot lie in the embedded Verma module, which proves the assertion about its lowest charge. \end{proof} \subsection{The conformal embedding} Let $S=\bigoplus_{j\geq0}L_j\subset\U$ be the subspace in Lemma~\ref{A:lat:blocklemma}. An intertwining operator between two $C_1$-cofinite modules has $C_1$-cofinite image when the target is lower-bounded \cite[Theorem 4.1.7]{A:CJORY}. Applying this to the lattice products, and using generic semisimplicity, shows that $S$ is closed under all vertex products. It contains $\one,T_\lambda$ and is a CFT-type sub-VOA. Normalize $W=W_1$ so that its $e^{-\beta}$ coefficient is one. The lowest-charge term of \[ W_{-2nj-1}W_j \] is a nonzero multiple of $e^{-(j+1)\beta}$. Since its weight is $h_{j+1}$, it has nonzero projection onto $L_{j+1}$. Thus $T_\lambda,W$ generate $S$. The singular $WW$ OPE lies in $L_0\oplus L_1$, because $h_2=6n>4n-1$. This also proves strong generation. \paragraph{The spin-one recoupling coefficients.} For generic $q$, put \[ [a]_q=\frac{q^a-q^{-a}}{q-q^{-1}},\qquad [a]_q!=\prod_{b=1}^a[b]_q,\qquad \Delta_j=\frac{[2-j]_q![j]_q!^2}{[j+3]_q!}. \] For $0\leq j,k\leq2$, the rational Racah matrix is \[ F_{jk}=[2k+1]_q\Delta_j\Delta_k \sum_{z=\max(2+j,2+k)}^{\min(4,2+j+k)} \frac{(-1)^z[z+1]_q!}{[z-2-j]_q!^2[z-2-k]_q!^2[4-z]_q![2+j+k-z]_q!^2}. \] Every factorial argument is nonnegative. Rational simplification gives $F^2=I$ and $F_{0r}=(-1)^r[2r+1]_q/[3]_q\ne0$. In particular a diagonal matrix commuting with $F$ is scalar. These are the exact recoupling facts used next. \begin{lemma}\label{A:lat:coupling} The vacuum coefficient and the $W$ self-coupling in $WW$ are both nonzero for $\lambda\ne0$. \end{lemma} \begin{proof} The charge-$-1$ term of $W$ gives a coefficient one of $\one$ in $o(W)\varepsilon$. Since $W$ is primary of even weight, its zero mode is self-adjoint. Consequently \eqref{A:lat:ell} and $o(W)\one=0$ give \begin{equation}\label{A:lat:zero} o(W)\varepsilon=\one+2\lambda\varepsilon. \end{equation} The Zhu product $[W]*[W]$ is a sum of descendants in $L_0\oplus L_1$: even its highest contributing weight is $4n<6n$. If $\nu$ and $\mu$ denote the vacuum coefficient and self-coupling, it has the form \[ y^2=\nu f_n(x)+\mu g_n(x)y,\qquad f_n(0)=0. \] Here $x=[T_\lambda]$, $y=[W]$; $f_n(0)=0$ follows because every contributing vacuum descendant has positive weight. Applying this relation to $\U_0$ gives $\mu g_n(0)=2\lambda$, so $\mu\ne0$. To determine the vacuum coefficient, use the integer-spin factorization in Section~\ref{sec:diagonal}. Compare multiplication with the canonical diagonal algebra in $\mathcal E\boxtimes\mathcal E^{\mathrm{rev}}$, where $\mathcal E=SO(3)_q$ and $q$ is not a root of unity. Normalize every channel relative to that canonical multiplication, and write its relative scalar as $a_{ij}^k$. Commutativity gives $a_{ij}^k=a_{ji}^k$. For three spin-one inputs and spin-one output, the two intermediate labels in the product category must coincide. In bases in the second factor dual to the bases in the first, the coefficient matrix is diagonal: \[ D=\operatorname{diag}\big(a_{11}^0,(a_{11}^1)^2,a_{11}^2a_{21}^1\big). \] The canonical diagonal multiplication is the identity pairing between the two spaces of three-point compositions. Therefore changing their parenthesization acts on these two spaces by $F$ and its inverse transpose. Associativity is exactly $FDF^{-1}=D$. In a rational Racah gauge, \[ F_{0r}=(-1)^r\frac{[2r+1]_q}{[3]_q},\qquad 0\leq r\leq2. \] All three entries are nonzero. Comparing entries $(0,r)$ in $FD=DF$ makes $D$ scalar. Since $a_{11}^1\ne0$, it follows that $a_{11}^0\ne0$, and hence $\nu\ne0$. The displayed Racah coefficients give the required nonzero matrix entries explicitly. \end{proof} The restricted invariant form on $S$ has radical equal to its maximal proper ideal. In its simple quotient, $W$ survives by Lemma~\ref{A:lat:coupling}. This quotient is a simple commutative diagonal algebra in the same product category. The support theorem for such algebras \cite[Proposition 4.4]{A:CKM} says that the surviving labels form a tensor subcategory. Spin one generates all integer spins. Hence every $L_j$ survives, the radical is zero, and $S$ is simple. Finally, the same recoupling calculation gives \[ \frac{\nu}{\nu_{\rm can}}= \left(\frac{\mu}{\mu_{\rm can}}\right)^2. \] After rescaling the canonical generator, the singular generator OPEs of $S$ and $V$ agree: the relevant Virasoro intertwining spaces are one-dimensional. Both are simple, CFT-type, and strongly generated by these generators. The common universal vacuum vertex algebra with these generator relations has a unique maximal proper graded ideal, so their simple quotients agree. This proves $S\cong V$ and the embedding in Theorem~\ref{A:lat:main}. \subsection{Simplicity of the complement} The nondegenerate form on $V$ gives a $V$-module orthogonal decomposition \[ \U=V\oplus X. \] The form on $X$ is nondegenerate. Distinct Virasoro blocks are orthogonal, and $X=\bigoplus X_j$ with composition factors $L_j,K_j,K_j$. There is no $L_j$ submodule of $X_j$, by the uniqueness in Lemma~\ref{A:lat:blocklemma}. \begin{lemma}\label{A:lat:essential} The socle of $X_j$ is one copy of $K_j$, and every nonzero Virasoro submodule of $X_j$ contains it. This copy is the singular submodule of the embedded $M_j$, and has lowest charge $j+1$. \end{lemma} \begin{proof} Only $K_j$ can occur in the socle. If two copies occurred there, self-duality would put two copies in the head as well. The kernel of the quotient onto that head would be a copy of $L_j$, contradicting the absence of an $L_j$ submodule. The socle is therefore a single $K_j$. Every nonzero submodule of a finite-length module contains a simple submodule. The singular copy in $M_j$ already provides this $K_j$ and fixes its lowest charge. \end{proof} Write $\phi_j$ for a highest vector of this socle. For $j\geq1$, the lowest-charge part of $Y(W,z)\phi_j$ is \begin{equation}\label{A:lat:lowering} Y(e^{-\beta},z)\phi_j^{(0)}\in\mathsf F_j\{z\}, \end{equation} where $\phi_j^{(0)}$ is the nonzero Feigin--Fuchs singular vector in $\mathsf F_{j+1}$. This series is nonzero: the exponential lattice field is a product of invertible creation and annihilation exponentials and a charge shift. Its image in the simple quotient $L_{j-1}$ of $\mathsf F_j\cong M_{j-1}$ is zero. Indeed, a nonzero intertwiner of type $\binom{L_{j-1}}{L_1\ K_j}$ would, by contragredient symmetry, give a nonzero intertwiner of type $\binom{K_j}{L_1\ L_{j-1}}$. The latter is impossible by the $C_1$ image theorem, since $K_j$ is not $C_1$-cofinite. Thus \eqref{A:lat:lowering} has nonzero image in $K_{j-1}\subset\mathsf F_j$. Exact block projection now shows that some coefficient of $Y(W,z)\phi_j$ has a nonzero component in $X_{j-1}$. By Lemma~\ref{A:lat:essential}, any $V$-submodule containing $\phi_j$ consequently contains $\phi_{j-1}$. Repeating this reaches \[ \phi_0=D\varepsilon. \] Equation \eqref{A:lat:zero} and translation covariance give \begin{equation}\label{A:lat:x0} W_{2n}D\varepsilon=2n(\one+2\lambda\varepsilon)=2n x_0, \qquad x_0:=\one+2\lambda\varepsilon\in X_0. \end{equation} The weight-zero vector $x_0$ satisfies \[ \langle x_0,x_0\rangle =\ell(\one+4\lambda\varepsilon)=-1, \] because $\varepsilon_{-1}\varepsilon=0$. Every nonzero $V$-submodule of $X$ contains some $\phi_j$, by block projection and Lemma~\ref{A:lat:essential}; hence it contains $x_0$. The submodule generated by $x_0$ is therefore simple. Its invariant form is nonzero and consequently nondegenerate. Its orthogonal complement in $X$ would be another submodule and, if nonzero, would also contain $x_0$, an impossibility. Thus $X$ is simple. This argument proves simplicity for all $n$, without assuming rationality. \subsection{The half-lattice decomposition and the simple current} Consider $\mathcal M_n=\bigoplus_{k\in\Z}\mathsf F_{k+1/2}$. Put \[ u_j=n(j^2-\tfrac14),\quad v_j=u_j+4j^2=(n+4)j^2-\tfrac n4 \quad(j\geq1). \] The argument of Lemma~\ref{A:lat:blocklemma} applies with even Kac label $2j$. The corresponding two Fock quotients have charges $1/2-j$ and $1/2+j$. There is exactly one $C_1$ simple submodule $L(c,u_j)$ in their block. The charge-$1/2$ Fock quotient is a separate one-element block, the irreducible Verma module $L(c,-n/4)$. Thus the $C_1$ socle is precisely \[ J=\bigoplus_{j\geq1}L(c,u_j). \] It is stable under $V$, again by the $C_1$ image theorem. Its $j$th highest vector has lowest-charge term $e^{(1/2-j)\beta}$, and the lowest-charge lattice product with $W$ gives a nonzero transition from the $j$th to the $(j+1)$st summand. The lattice coset $\mathcal M_n$ is self-contragredient and has a nondegenerate invariant pairing: the lattice intertwiner $\mathcal M_n\times \mathcal M_n\to\U$, paired with the invariant form on $\U$, identifies $\mathcal M_n$ with its contragredient. Equivalently, this follows from the order-two lattice fusion rule. The next argument shows that the restriction of this pairing to $J$ is nondegenerate. The lattice intertwiner sends $J\times J$ into $V$, by $C_1$ closure. Let $R$ be the radical of the restricted pairing. It is a $V$-submodule. Invariance and skew-symmetry imply that $Y(R,z)J$ pairs to zero with all of $V$, hence is zero. But take the lowest vector $j_1$ of $J$, whose lowest-charge term is $e^{-\beta/2}$. For any nonzero $r\in R$, the lowest-charge part of $Y(j_1,z)r$ is a nonzero exponential field acting on the lowest-charge part of $r$. It cannot vanish. This contradiction proves nondegeneracy. Each Virasoro summand of $J$ now has nondegenerate pairing. Any $V$-submodule is a sum of these distinct simple summands and therefore has a $V$-stable orthogonal complement. A nonzero submodule contains all sufficiently large labels by the upward transition above. Two nonzero complementary submodules cannot both do this. Thus $J$ is simple. Consequently $\mathcal M_n=J\oplus Y$ orthogonally. In each two-element Virasoro block, $Y$ has one $L(c,u_j)$ and two $L(c,v_j)$ factors and has essential socle $L(c,v_j)$, by Lemma~\ref{A:lat:essential}. The same lowest-charge lowering argument sends this socle to the one with label $j-1$. At $j=1$ it reaches the isolated irreducible block $L(c,-n/4)$. Every nonzero submodule of $Y$ therefore contains that block. Its bottom space is the one-dimensional bottom space of $\mathcal M_n$ and has nonzero pairing. The argument used for $X$ proves that $Y$ is simple. The standard lattice abelian intertwining algebra $\U\oplus \mathcal M_n$ contains the closed subalgebra \[ E=V\oplus J. \] Closure follows from the $C_1$ image theorem for each pair of summands. Both summands are simple $V$-modules, and the pairing on $J$ makes the vacuum channel in $JJ$ nonzero. Hence $E$ is simple. The involution acting as $+1$ on $V$ and $-1$ on $J$ has fixed-point algebra $V$. For odd $n$, the preceding $C_2$ theorem supplies the full braided module category of $V$. McRae's abelian-intertwining-algebra orbifold theorem \cite[Corollary 4.8]{A:Orb} applies to $E$ and this involution, without assuming rigidity of that full category. It gives $J\boxtimes_VJ\cong V$. The inherited lattice associator and braiding give $F(J,J,J)=-1$ and $c_{J,J}=i^n$. Also $h_J=3n/4$, so $\theta_J=i^{-n}$, and the ribbon dimension on this rigid rank-two subcategory is $-1$. \subsection{The remaining lattice modules} Let $a\not\equiv0,n\pmod{2n}$ and $r=a/(2n)$. The Fock quotients of $\mathcal M_a$ have highest weights \[ h(k)=n(k+r)(k+r-1),\qquad k\in\Z. \] Each is an irreducible Verma module: a positive Kac label would force $2(k+r)$ to be an integer. They lie in distinct one-element Virasoro blocks. The charge filtration therefore gives a multiplicity-free direct sum of these irreducibles for the deformed Virasoro action. Denote the corresponding submodules by $Q_k$. The lowest-charge term of $W$ gives a nonzero transition $Q_k\to Q_{k-1}$, just as in \eqref{A:lat:lowering}. The lattice contragredient of $\mathcal M_a$ is $\mathcal M_{2n-a}$; its block paired with $Q_k$ has charge $1-r-k$. The downward transition on that dual module, together with primary-field adjointness, gives a nonzero upward transition $Q_k\to Q_{k+1}$. Every nonzero $V$-submodule consequently contains every $Q_k$. Thus the restriction is simple. The paired modules have the same lowest conformal weight and the same $W$ zero-mode eigenvalue, since $W$ has even weight. Their one-dimensional top spaces are therefore isomorphic as Zhu-algebra modules: the Zhu algebra is generated by $[T]$ and $[W]$. Uniqueness of the simple module with a specified simple top proves $\mathcal M_a\cong \mathcal M_{2n-a}$ upon restriction. Put $\Lambda_{b,m}=\theta_{b,m}/\eta$, with $\theta_{b,m}=\sum_{k\in\Z}q^{(b+2mk)^2/(4m)}$. The modules constructed above have characters \begin{center} \begin{tabular}{ccl} \toprule Module & Lowest weight & Character\\ \midrule $V$ & $0$ & $\frac12(\Lambda_{n,n}-\Lambda_{n+4,n+4})$\\[3pt] $X$ & $0$ & $\frac12(\Lambda_{n,n}+\Lambda_{n+4,n+4})$\\[3pt] $J$ & $3n/4$ & $\frac12(\Lambda_{0,n}-\Lambda_{0,n+4})$\\[3pt] $Y$ & $-n/4$ & $\frac12(\Lambda_{0,n}+\Lambda_{0,n+4})$\\[3pt] $\mathcal M_a$, $1\leq a2n. \] Hence all reduced terms lie in Virasoro summands with label at most $r$. The other terms in \eqref{A:rat:up} have classes polynomial in $x_l,x_r$ times $[j_s]$, $s\leq r$. Since $\gamma_r\ne0$, induction expresses every $[j_{r+1}]$ in terms of $[j]$. This proves cyclicity. Every term in $W*j$ or $j*W$ has weight at most $u_1+2n1$, an extension $0\to T\to E\to V\to0$ splits: lift the vacuum to $v\in E_0$. Then $L_{-1}v\in T_1=0$. A translation-invariant module vector is vacuum-like, so $a\mapsto a_{-1}v$ gives a module map $V\to E$ splitting the quotient. Nonintegral $h_T$ also gives a split extension by the $L_0$ residue decomposition, and $h_T=0$ is covered by Theorem~\ref{A:rat:main}. At this stage a possible obstruction to projectivity of $V$ must have lowest weight in the explicit finite arithmetic set \begin{equation}\label{A:rat:badset} \begin{split} \mathcal B_n=\{k\in\Z:\;&-\lfloor n/4\rfloor\leq k\leq1,\quad k\ne0,\\ &n(n+4k)\text{ or }(n+4)(n+4k)\text{ is a nonnegative integer square}\}. \end{split} \end{equation} There are at most two simple candidates at each such weight. The next argument removes all negative weights from this obstruction set, including at the odd-square parameters relevant to the Haagerup--Izumi family. \subsection{Extensions with negative lowest weight} Fix $t>1$ with $t+t^{-1}=n+2$, and use momentum notation \[ h(p)=\frac{p^2-(t-1)^2}{4t}. \] For rational labels $r,s$, the weight $h(rt-s)$ is rational precisely when $r^2=s^2$. This follows by reducing $t^{-1}=n+2-t$ and using irrationality of $t$. The level-nine null vector of $W=L(c,h_{3,3})$ gives the usual nine-root fusion restriction \begin{equation}\label{A:rat:Wroots} p\longmapsto p+2it-2j,\qquad i,j\in\{-1,0,1\}. \end{equation} To see the restriction explicitly, apply the Virasoro bimodule identity \eqref{A:rat:VirZhu} to the level-nine singular vector in Appendix~\ref{app:null}, with input weight $h$ and output weight $\widetilde h$. The resulting monic polynomial is \begin{equation}\label{A:rat:Wpoly} \begin{split} P_{3,3}(h,\widetilde h)={}&(\widetilde h-h) \prod_{a\in\{1,t,t-1,t+1\}}\left[ \left(\widetilde h-h-\frac{a^2}{t}\right)^2 -\frac{a^2\bigl(4th+(t-1)^2\bigr)}{t^2}\right]. \end{split} \end{equation} For $h=h(p)$, each quadratic factor has roots $h(p+2a)$ and $h(p-2a)$. Together with the root $h(p)$ these are exactly the nine weights in \eqref{A:rat:Wroots}. Every nonzero three-point coupling annihilates this polynomial, which gives the necessary fusion restriction used below. \begin{lemma}\label{A:rat:genericorbit} Suppose a simple ordinary $V$-module has lowest weight $h(k(t-1))$ or $h(k(t+1))$, where $k\in\Q\setminus\Z$. Its Virasoro constituents lie, respectively, in the sets \[ \{L(c,h((k+2m)(t-1))):m\in\Z\},\qquad \{L(c,h((k+2m)(t+1))):m\in\Z\}. \] All these constituents are irreducible Verma modules in singleton Virasoro blocks. In particular, none lies in a block containing a constituent of $V$. \end{lemma} \begin{proof} For the first family, a shift in \eqref{A:rat:Wroots} changes the rational labels to $r=k+2i$, $s=k+2j$. Rational output weight requires \[ (k+2i)^2-(k+2j)^2=4(i-j)(k+i+j)=0. \] Since $k\notin\Z$, this forces $i=j$, giving the stated step-two orbit. For the second family the corresponding equation is \[ (k+2i)^2-(k-2j)^2=4(i+j)(k+i-j)=0, \] so $i=-j$. These nonintegral labels cannot be integer Kac labels, even after a momentum sign change. Feigin--Fuchs linkage therefore makes each orbit weight a singleton block \cite[Theorem 3.1]{A:BNW}. To propagate the restriction through the module, restrict the ordinary module to Virasoro and apply Lemma~\ref{A:blocks:union} to extend the category-$\mathcal O$ block projections. The hypotheses here include ordinary $L_0$-action; generalized modules are not being projected by this argument. A fixed singleton or two-element block has finite length, since its possible composition-factor lowest spaces are finite-dimensional. If the projection of $Y(W,z)$ from a simple constituent to a block is nonzero, first take the Virasoro submodule spanned by its coefficients and then a simple quotient of that image. The block has finite length by the preceding weight-space bound, so such a quotient exists and the composed intertwiner is nonzero. Its output weight must satisfy \eqref{A:rat:Wroots}. Any permitted output belongs to a singleton block in the displayed orbit; hence a block outside that orbit cannot receive a nonzero image. Thus no block outside the displayed orbit is reached. Singleton ordinary blocks are semisimple. Iterating from the lowest vector and using strong generation by $T,W$ proves the claim. The blocks of $V$ have positive integral diagonal Kac labels and their linked negative-label partners, so they are disjoint from these orbits. \end{proof} \begin{proposition}\label{A:rat:negativeext} For every simple $T$ with negative lowest weight, \[ \Ext^1(V,T)=0. \] Consequently the only remaining possible obstruction to projectivity of $V$ is a simple module of lowest weight one. \end{proposition} \begin{proof} Nonintegral lowest weights split by the $L_0$ residue decomposition. If $h_T<0$ is integral, the spectral theorem gives either $k=a/n$ or $k=a/(n+4)$ with $0<|k|<1$. The zero label would give $h_T=-n/4$, which is not integral for odd $n$. Lemma~\ref{A:rat:genericorbit} applies. In an extension $0\to T\to E\to V\to0$, the nilpotent part of $L_0$ factors through a homomorphism $V\to T$, and is therefore zero. Thus $E$ is ordinary, so no generalized-category-$\mathcal O$ assertion is being imported. The disjoint Virasoro blocks split the sequence as Virasoro modules. The section with the Virasoro constituents of $V$ is stable under $W$: fusion of the $C_1$-cofinite $W$ with the diagonal constituents of $V$ has $C_1$-cofinite image, and cannot reach any of the singleton generic blocks of $T$ \cite{A:CJORY}. Hence this section is a $V$-submodule and splits the extension. Combining this with the previous $h_T>1$, nonintegral and $h_T=0$ arguments proves the last statement. \end{proof} \subsection{The first weight-one candidate} Let $x_0\in X_0$, $y_0\in Y_{-n/4}$ be nonzero. The lattice construction gives \[ x_0=\one+2\lambda\varepsilon,\qquad L_{-1}x_0\ne0,\qquad (x_0,x_0)\ne0, \qquad o(W)x_0=w_Xx_0,\quad w_X\ne0. \] The vector $\phi=L_{-1}x_0$ is a Virasoro highest vector of weight one. Use the nonzero intertwiner $\mathcal Y$ of type $\binom X{J\ Y}$ and the invariant form on $X$. For homogeneous $v\in J$ of weight $b$, write $T_s(v)$ for the coefficient of $z^{s+n/4-b}$ in $\mathcal Y(v,z)y_0$, and set \[ \Phi_0(v)=(x_0,T_0(v)),\qquad \Phi_1(v)=(\phi,T_1(v)). \] The first functional is the actual lowest-space Zhu map, and $\Phi_0(j)\ne0$. The $L_1$ bracket identity gives \begin{equation}\label{A:rat:Phi} \Phi_1(v)=\Phi_0(L_1v)+(b+n/4)\Phi_0(v), \qquad \Phi_1(j)=n\Phi_0(j)\ne0. \end{equation} The second functional is a Virasoro highest-to-highest functional with ordered weights $(h,h')=(-n/4,1)$; it is not a lowest-$V$-space map. Put $w=2n$ and $C=W\circ j=\sum_{i=0}^w\binom wi W_{i-2}j$. Since $\Phi_0(C)=0$, \eqref{A:rat:Phi} and \[ [L_1,W_{i-2}]=(2w-i)W_{i-1} \] give, by shifting one binomial sum, \begin{align*} \Phi_1(C) &=\sum_{i=0}^w\binom wi \bigl((2w-i)\Phi_0(W_{i-1}j)-i\Phi_0(W_{i-2}j)\bigr)\\ &=w\sum_{i=0}^w\binom wi\Phi_0(W_{i-1}j) =2n\,w_X\Phi_0(j). \end{align*} For $n\geq3$, all terms of $C$ lie below the next Virasoro primary of $J$. Evaluating their first-channel reduction on $\Phi_1$ therefore proves the exact identity \begin{equation}\label{A:rat:lowerexclusion} q\,R_{n,1}(1,-n/4)=2w_X\ne0. \end{equation} \begin{corollary}\label{A:rat:onepair} For odd $n\geq3$, the module $Y$ is the unique simple of lowest weight $-n/4$. There is at most one simple module $T$ of lowest weight one, and, if it exists, it satisfies \[ h_{J\bo T}=\frac{3n}{4}+4. \] \end{corollary} \begin{proof} At $h=-n/4$, the spectral theorem permits partners of weight $0$ or $1$. Identity \eqref{A:rat:lowerexclusion} excludes the second; ordered-pair uniqueness identifies the first with $Y$. At $h=1$ the first quadratic branch is impossible because $n(n+4)$ is not a square. The two second-branch choices have partners $-n/4$ and $3n/4+4$. The former is excluded by applying the involution $J\bo-$ to \eqref{A:rat:lowerexclusion}. Ordered-pair uniqueness gives the remaining at-most-one assertion. \end{proof} \subsection{The second weight-one candidate} Specialize the fourth-order operator to \[ x_r=1,\qquad x_l=\frac{3n}{4}+4. \] For its scalar solution $F(z)=z^{-2n}\sum_{\ell\geq0}f_\ell z^\ell$, set \[ P(z)=z^{2n}(1+z)^{2n}F(z). \] After this gauge transformation and division by $1+z$, the equation is \begin{equation}\label{A:rat:gauged} \mathcal G_n P=0,\qquad \mathcal G_n=\sum_{k=0}^3 z^k C_k(\theta),\qquad \theta=z\partial_z, \end{equation} where the following four polynomials specify the operator completely: \begin{align*} C_0(e)={}&e(e-3n)\bigl(e^2-(3n+2)e-6n+1\bigr),\\ C_1(e)={}&3e^4-(22n+14)e^3+(51n^2+54n+13)e^2\\ &+(-36n^3-18n^2+9n-6)e-54n^3-90n^2+36n,\\ C_2(e)={}&3e^4-(26n+22)e^3+(79n^2+126n+45)e^2\\ &-(96n^3+206n^2+129n+14)e+36n^4+78n^3+36n^2-32n,\\ C_3(e)={}&(e-2n)(e-3n-4) \bigl(e^2-(5n+6)e+6n^2+14n+9\bigr). \end{align*} The recurrence \eqref{A:rat:Dr} gives these coefficients as polynomial identities in $n$. \begin{lemma}\label{A:rat:twocircle} For every odd integer $n\geq3$, \[ \bigl(R_{n,1}(3n/4+4,1),R_{n,2}(3n/4+4,1)\bigr)\ne(0,0). \] Consequently there is no simple $V$-module of lowest weight one. \end{lemma} \begin{proof} Suppose both circle coefficients vanish. The recurrence determines $f_\ell$ uniquely through $\ell=2n+2$, because those levels precede $3n$. Put \[ p_\ell=\sum_{i=0}^{\min(2n,\ell)}\binom{2n}{i}f_{\ell-i}, \qquad P(z)=\sum_{\ell=0}^{2n}p_\ell z^\ell. \] Then $P(0)=1$ and $\deg P\leq2n$. The assumed circle relations say $p_{2n+1}=p_{2n+2}=0$. The gauged recurrence therefore makes every coefficient of $\mathcal G_nP$ through degree $2n+2$ vanish. Its only possible further coefficient is at degree $2n+3$, and this is $C_3(2n)p_{2n}=0$. Thus $\mathcal G_nP=0$. This step uses no coefficient at or beyond the resonant level $3n$. The same equation has an explicit polynomial solution \begin{equation}\label{A:rat:highsol} S_n(z)=z^{3n}H_n(z), \end{equation} where \begin{equation}\label{A:rat:Hpoly} \begin{split} H_n(z)={}&1+2z+\frac{2(4n-1)}{6n-1}z^2\\ &+\frac{2(2n-1)(5n-4)}{3(3n-4)(6n-1)}z^3 +\frac{n(2n-1)}{3(3n-4)(6n-1)}z^4. \end{split} \end{equation} Substitution in \eqref{A:rat:gauged} verifies this identity coefficient by coefficient. In particular, \begin{equation}\label{A:rat:Hminusone} \deg S_n=3n+4,\qquad H_n(0)=1,\qquad H_n(-1)=-\frac{4(2n-1)}{3(3n-4)(6n-1)}\ne0. \end{equation} At $z=-1$, the lowest-order coefficient of $\mathcal G_n(z+1)^r$ is the indicial polynomial \begin{equation}\label{A:rat:indminusone} I_{-1}(r)=r(r-2n-8)\bigl(r^2-(2n+6)r+2n+9\bigr). \end{equation} The two roots of its quadratic factor are $n+3\pm\sqrt{n(n+4)}$, and are irrational. It follows directly by examining the first nonzero Taylor coefficient that any nonzero polynomial solution vanishing at $-1$ has vanishing order exactly $2n+8$. No connection formula or absence-of-logarithms assumption is needed for this assertion. Now form the polynomial Wronskian \[ \mathscr W=P S_n'-P'S_n. \] At zero, $P(0)=1$ and $S_n=z^{3n}(1+O(z))$, so \[ \operatorname{ord}_0\mathscr W=3n-1; \] in particular, $\mathscr W\ne0$. Since $S_n(-1)\ne0$, the nonzero polynomial solution \[ Q=P-\frac{P(-1)}{S_n(-1)}S_n \] vanishes at $-1$. It is nonzero because $Q(0)=1$. By \eqref{A:rat:indminusone}, $\operatorname{ord}_{-1}Q=2n+8$. Replacing $P$ by $Q$ leaves $\mathscr W$ unchanged, so \[ \operatorname{ord}_{-1}\mathscr W=2n+7. \] Consequently this nonzero polynomial must have degree at least \[ (3n-1)+(2n+7)=5n+6. \] But its definition gives \[ \deg\mathscr W\leq\deg P+\deg S_n-1\leq5n+3, \] a contradiction. The two circle coefficients cannot both vanish. Corollary~\ref{A:rat:onepair} and the Zhu relations \eqref{A:rat:circles} therefore exclude the last weight-one candidate. \end{proof} \subsection{Projectivity of the vacuum and rationality} \begin{lemma}[Projective unit criterion]\label{A:rat:projectiveunit} Let $\mathcal D$ be a finite $\C$-linear abelian braided $r$-category with simple unit and right-exact tensor product. If its unit is projective, then $\mathcal D$ is rigid and semisimple. \end{lemma} \begin{proof} Here an $r$-category is a Grothendieck--Verdier category whose dualizing object is the tensor unit; write $D$ for its contravariant duality. For a nonzero indecomposable $M$, the evaluation $DM\otimes M\to\one$ corresponds to $\id_{DM}$ under the defining adjunction and is therefore nonzero. Since the unit is simple, this evaluation is an epimorphism. Projectivity of $\one$ splits it, so $M$ is non-negligible in the exact sense of Etingof--Penneys \cite[Theorem 1.1]{A:EP}. Moderate growth is automatic here and does not require already knowing rigidity. Let $K$ be the maximum of $1$ and the lengths of $S_i\otimes S_j$ over the finitely many simple objects. Right exactness in each variable gives \[ \ell(A\otimes B)\leq K\ell(A)\ell(B),\qquad \ell(M^{\otimes r})\leq K^{r-1}\ell(M)^r. \] Also $\dim\operatorname{Hom}(A,B)\leq\ell(A)\ell(B)$ by induction on composition series. Thus $\dim\operatorname{End}(M^{\otimes r})0. \] Then \[ \FP(M)=\frac{S_{MY}}{S_{VY}} =\lim_{\epsilon\downarrow0}\frac{\chi_M(i\epsilon)}{\chi_V(i\epsilon)} \] for every simple $M$. \end{lemma} \begin{proof} Modular transformation and uniqueness of $Y$ show that the coefficients $v_M=S_{MY}$ of the fastest exponential are real and nonnegative, since the characters have nonnegative graded dimensions. By hypothesis $v_V>0$. Verlinde diagonalization says that $v$ is a simultaneous eigenvector of the nonnegative fusion matrices $N_M$, with eigenvalue $\lambda_M=v_M/v_V\geq0$. Since $M\otimes M^*$ contains the unit, $N_MN_{M^*}\geq I$ entrywise. Applying this to the nonzero vector $v\geq0$ yields $\lambda_M\lambda_{M^*}\geq1$. Thus all $\lambda_M$ and all entries of $v$ are strictly positive. A nonnegative matrix with a strictly positive eigenvector has that eigenvalue as its spectral radius. Therefore $\lambda_M=\FP(M)$, and the asserted limit follows. \end{proof} Poisson summation and the eta transformation give \[ \Lambda_{a,m}(i\epsilon)\sim\frac{1}{\sqrt{2m}} e^{\pi/(12\epsilon)}. \] By \eqref{A:cat:characters}, set \begin{equation}\label{A:cat:tdef} t=\FP(X)=\frac{(2n)^{-1/2}+(2n+8)^{-1/2}} {(2n)^{-1/2}-(2n+8)^{-1/2}} =\frac{n+2+\sqrt{n(n+4)}}2. \end{equation} The denominator is positive. Useful identities are \begin{equation}\label{A:cat:tidentities} t+t^{-1}=n+2,\qquad (t-1)^2=nt,\qquad \FP(A)=1+t. \end{equation} \subsection{The near-group fusion rules}\label{A:cat:neargroup} The dimension formulas for connected \'etale algebras \cite[Lemma 3.11 and Corollary 3.32]{A:DMNO} imply \begin{equation}\label{A:cat:globalFP} \FP(\D_n)=n(1+t)^2,\qquad \FP(\Ncat_n)=n(1+t). \end{equation} Let $r=\FP(R)$. Subtracting the $n$ known invertible simples gives \[ r^2=n(1+t)-n=nt=(t-1)^2. \] Therefore $r=t-1>1$, so $R$ is noninvertible. The local invertibles permute the noninvertible simples on both sides; there is just one. Hence $gR=Rg=R$ and $R^*=R$. Frobenius reciprocity makes every invertible occur once in $R^2$. Write \[ R^2=\bigoplus_g g\oplus mR. \] Dimensions now give \[ m=\frac{r^2-n}{r}=\frac{n(t-1)}{t-1}=n. \] This proves \eqref{A:cat:NG}. An exact induction decomposition will also be needed. Adjunction and \eqref{A:cat:branch} show that $F_A(X)$ contains the unit once and no other local invertible. Thus $F_A(X)=\one\oplus bR$. Since the tensor functor $F_A$ preserves Frobenius--Perron dimensions, $t=1+b(t-1)$, and \begin{equation}\label{A:cat:indX} F_A(X)=\one\oplus R. \end{equation} \subsection{Spherical dimensions}\label{A:cat:spherical} Write $x=d_{\D_n}(X)$. The free-module functor is pivotal. By \eqref{A:cat:indX}, $x=1+d(R)$, while \eqref{A:cat:NG} and the fact that odd-order invertibles have spherical dimension one give \[ d(R)^2=n+n\,d(R). \] Consequently \[ x\in\{t,t^{-1}\},\qquad x>0. \] The character asymptotics determine the root. By (H5) and \eqref{A:cat:characters}, \begin{equation}\label{A:cat:a0} a:=S_{VY}=\frac12\left(\frac1{\sqrt{2n}}-\frac1{\sqrt{2(n+4)}}\right)>0. \end{equation} Because $Y=JX$ and $d(J)=-1$, $d(Y)=-x$. The categorical dimension formula for local modules gives \[ \dim\D_n=n(1+x)^2,\qquad \dim\mathcal C_n=2n(1+x)^2. \] The normalized VOA $S$-matrix satisfies $S_{VY}/S_{VV}=d(Y)$ and $S_{VV}^2=1/\dim\mathcal C_n$. Squaring eliminates any choice of sign in its normalization. Since $x>0$, \eqref{A:cat:a0} gives \[ \frac{x}{1+x}=\sqrt{2n}\,a =\frac12\left(1-\sqrt{\frac n{n+4}}\right). \] Solving yields \begin{equation}\label{A:cat:negdim} d(X)=t^{-1},\qquad d(R)=t^{-1}-1=\frac{n-\sqrt{n(n+4)}}2. \end{equation} In particular the spherical near-group category obtained from the VOA has negative noninvertible dimension. Frobenius--Perron dimensions remain the positive numbers computed above. \subsection{Condensation at a square parameter}\label{A:cat:condensation} Now put $n=N^2$ with $N$ odd, and let $H=N\Z/N^2$. In $\Pcat_{N^2}^{\mathrm{rev}}$ its twists and braiding are trivial. Also \[ H^\perp=H, \] because the associated bicharacter is $e^{-4\pi ixy/N^2}$ and $2$ is invertible modulo $N$. Hence the central subcategory on $H$ is Tannakian and Lagrangian in this pointed factor. Let $B=\bigoplus_{h\in H}h$ be its regular central algebra in $Z(\Ncat_{N^2})\simeq\D_{N^2}\bo\Pcat_{N^2}^{\mathrm{rev}}$, and define \begin{equation}\label{A:cat:Fdef} \Fcat_N:=\operatorname{Rep}_{\Ncat_{N^2}}(B). \end{equation} The forgetful functor identifies this central pointed subcategory with the invertibles $H\subset\Ncat_{N^2}$, so this is ordinary de-equivariantization by $\widehat H$. Its unit is simple. The local modules of the second factor over $B$ are $\mathrm{Vec}$. The center theorem for a central algebra \cite[Theorem 3.20]{A:DMNO} therefore gives \begin{equation}\label{A:cat:centerF} Z(\Fcat_N)\simeq Z(\Ncat_{N^2})_B^{\mathrm{loc}} \simeq\D_{N^2}. \end{equation} There is also a concrete description directly in VOA modules. Put $\alpha=\beta/N$, so $\alpha^2=2$, and use the same conformal vector in the overlattice extension \[ B_N=(V_{\Z\alpha},T_\lambda). \] Condensation in stages identifies \eqref{A:cat:Fdef} with \begin{equation}\label{A:cat:Fconcrete} \Fcat_N\simeq(\D_{N^2})_{B_N},\qquad B_N\cong\one\oplus X\oplus 2\bigoplus_{r=1}^{(N-1)/2} Z_{Nr}. \end{equation} The regular algebra on $H$ is connected \'etale in the local $A$-module category, so its algebra over $A$, namely $B_N$, is connected \'etale in $\D_{N^2}$ \cite[Proposition 3.16]{A:DMNO}. Its twist is trivial. All simple $Z_{Nr}$ displayed here are distinct. The grading of this overlattice VOA can contain negative weights; no CFT-type extension theorem is used for it. Each invertible local $A$-module has spherical dimension one, so its restriction has dimension $d(A)=1+t^{-1}$. The regular algebra on $H$ has $N$ such summands. Consequently \[ d_{\D_{N^2}}(B_N)=N\,d_{\D_{N^2}}(A)=N(1+t^{-1})\ne0. \] Together with $\theta_{B_N}=\id$, this verifies the dimension hypothesis for the spherical module structure of \cite{A:KO} on \eqref{A:cat:Fconcrete}. Its central free-module functor is pivotal. Thus the braided equivalence \eqref{A:cat:centerF} also preserves twists: it is a ribbon equivalence with the inherited spherical center. As an independent rank check, the Lagrangian algebra $B_N$ corresponds to $I_{\Fcat_N}(\one)$, and hence \[ \rank\Fcat_N=\dim\End_{\D_{N^2}}(B_N) =1+1+4\frac{N-1}{2}=2N. \] \subsection{The Haagerup--Izumi fusion rules}\label{A:cat:HIfusion} The proof first determines the simple objects, then their left and right actions and their products. \textbf{Simple objects.} The free modules $gB$ on invertibles are indexed by $G/H=\Z/N$ and give precisely $N$ invertibles $\alpha_a$. By adjunction, \[ \End_B(RB)\cong\bigoplus_{h\in H}\Hom(R,Rh). \] Every summand is one-dimensional. Compositions are nonzero scalar multiples of the group law, so this algebra is a twisted group algebra of $H$. A cyclic group has $H^2(H,\C^\times)=0$; thus it is $\C^N$. Consequently $RB$ is a multiplicity-one sum of $N$ distinct simple objects $Q_i$. None is an $\alpha_a$, by adjunction. Induction is dominant, so these $2N$ objects exhaust the simples. Put $\delta=(N+\sqrt{N^2+4})/2$, so $\delta^2=1+N\delta$ and $\FP(R)=N\delta$. If $d_i=\FP(Q_i)$, tensor induction and de-equivariantization give \[ \sum_i d_i=N\delta,\qquad \sum_i d_i^2=\frac{N^2+(N\delta)^2}{N}-N=N\delta^2. \] Equality in Cauchy--Schwarz implies $d_i=\delta$ for every $i$. They are therefore noninvertible. \textbf{Free left and right actions.} For a noninvertible simple $Q$, write the total multiplicities of invertible and noninvertible terms in $QQ^*$ as $a,b$. Dimensions give \[ 1+N\delta=a+b\delta. \] As $\delta$ is irrational, $a=1$ and $b=N$. Frobenius reciprocity identifies $a$ with the order of the left stabilizer of $Q$. Thus the left action of $\Z/N$ is free and, since there are $N$ noninvertibles, transitive. The same argument using $Q^*Q$ proves the right action free and transitive. Choose $\rho=Q_0$ and write the others as $\alpha_a\rho$. There is an automorphism $\varphi$ of $\Z/N$ with \[ \rho\alpha_a=\alpha_{\varphi(a)}\rho. \] \textbf{The dual action supplies all translations.} De-equivariantization gives $\Ncat_{N^2}\simeq\Fcat_N^{\widehat H}$. The dual action fixes every invertible $gB$: twisting the $B$-action is undone by rescaling the $H$-graded free factor. Noninvertible orbits each produce at least one noninvertible in the equivariantization. Since $\Ncat_{N^2}$ has just one, the $N$ noninvertibles of $\Fcat_N$ form a single orbit. Thus $\widehat H$, of order $N$, acts freely and transitively on them. It fixes the $\alpha_a$ and commutes with their left action, so on simple labels it acts by every translation \[ \alpha_a\rho\longmapsto\alpha_{a+t}\rho. \] \textbf{Inversion and self-duality.} The dimension argument above applies also to $\rho^2$, so it has a unique invertible summand, say $\alpha_\ell$. Every translation fixes that summand, but \[ (\alpha_t\rho)^2=\alpha_{t+\varphi(t)}\rho^2. \] Comparing the unique invertibles shows $t+\varphi(t)=0$ for every $t$; hence $\varphi=-1$. Write $\rho^*=\alpha_j\rho$. Dualizing again gives $\rho=\alpha_{2j}\rho$. Freeness and oddness force $j=0$. Thus $\rho^*=\rho$, and the invertible summand of $\rho^2$ is $\one$. \textbf{Multiplicity one.} Write \[ \rho^2=\one\oplus\bigoplus_a m_a\alpha_a\rho,\qquad\sum_a m_a=N. \] Since $(\alpha_t\rho)^2=\rho^2$, applying each translation automorphism gives $m_{a+t}=m_a$. Therefore $m_a=1$ for all $a$. This proves all the rules \eqref{A:cat:HI}. The spherical de-equivariantization is pivotal, so $d(R)=\sum_i d(Q_i)$. All $\alpha_a$ have dimension one (they have odd order), and the $Q_i=\alpha_a\rho$ have a common dimension. Equation \eqref{A:cat:negdim} now yields \[ d(\rho)=\frac{d(R)}N=\frac{N-\sqrt{N^2+4}}2, \] as asserted. \subsection{The pointed associator}\label{A:cat:associator} Work first with $G=\Z/N^2$, $H=N\Z/N^2$, and central half-braiding \[ \sigma_h(x)=e^{2\pi i k hx/N^2}\quad(h\in H,x\in G). \] The pointed category before condensation has trivial associator. Choose coset representatives $0\leq a1$ in \eqref{A:cat:tdef}. Projection of associativity on $X^3$ to the $X$ output gives \[ aU+b^2V=0,\qquad V=f(f\otimes\id)-f(\id\otimes f). \] Thus $b\ne0$. If $(a_0,b_0)$ are the constants of the existing lattice multiplication, the same equation and $U\ne0$ give $a/b^2=a_0/b_0^2$. Set $s=b/b_0$. The identities \[ a=s^2a_0,\qquad sb=s^2b_0 \] are exactly the two components of $\varphi m=m_0(\varphi\otimes\varphi)$ for $\varphi=\id_{\one}\oplus s\id_X$. Thus $\varphi$ is an algebra isomorphism to the lattice algebra. The VOA-extension correspondence transports this isomorphism to an isomorphism fixing $V$ \cite[Theorem 3.2 and Remark 3.3]{B:HKL}. The extension has a two-dimensional weight-zero space; Remark~3.3 supplies the version of the correspondence that allows this. \section{The simple modules}\label{sec:spectrum} The next step determines the simple objects of $\D=\D_n$. Write $F:\D\to\Ncat_n$ for the free-module functor. The reconstruction gives \begin{equation}\label{B:FX} F(X)=\one\oplus R,\qquad t:=\FP(X)=\frac{n+2+\sqrt{n(n+4)}}2,\qquad \FP(R)=t-1, \end{equation} and \begin{equation}\label{B:dims} \begin{gathered} \delta:=d(R)=\frac{n-\sqrt{n(n+4)}}2,\qquad d(X)=1+\delta=t^{-1},\\ \dim\D=n(2+\delta)^2,\qquad \FP(\D)=n(1+t)^2. \end{gathered} \end{equation} The known objects are $\one,X$ and $Z_x$, $1\leq x\leq(n-1)/2$. The unknown induction multiplicities are first expressed categorically, then bounded using the lattice structure of $X$. \subsection{Induction multiplicities} \begin{lemma}\label{B:multiplicities} Write $Q_i$ for the as-yet-unclassified simples of $\D$, with $i$ in a finite set $I$. Then \begin{align} F(Z_x)&=x\oplus(-x)\oplus R, & F(Q_i)&=b_iR,\\ \FP(Q_i)&=b_i(t-1), & d(Q_i)&=b_i\delta, \end{align} where $b_i$ are positive integers, and \begin{equation}\label{B:squares} \sum_{i\in I}b_i^2=\frac{n+3}{2}=:M. \end{equation} Also \begin{equation}\label{B:XX} X^2=\one\oplus X\oplus\bigoplus_{x=1}^{(n-1)/2}Z_x \oplus\bigoplus_{i\in I} b_iQ_i, \qquad N_{X,Q_i}^{Q_i}=b_i^2-1. \end{equation} \end{lemma} \begin{proof} Adjunction with $U(g)$ determines every invertible summand of a free module. The known $Z_x$ have exactly the two invertible summands $x,-x$; an unlisted simple has none. Dominance gives $b_i>0$. The dimension formula for restriction of an $A$-module gives $\FP(Z_x)=\FP(A)=1+t$, since its local module has dimension one. Since $\FP(R)=t-1$, this proves the displayed decomposition of $F(Z_x)$. Pivotality gives the spherical dimensions. Subtract the dimensions of the known simples from $\FP(\D)=n(1+t)^2$. Using $(t-1)^2=nt$ gives \[ \sum_i b_i^2= \frac{n(1+t)^2-1-t^2-\frac{n-1}{2}(1+t)^2}{(t-1)^2} =\frac{n+3}{2}. \] Adjunction also gives $U(R)=X\oplus\bigoplus_x Z_x\oplus\bigoplus_i b_iQ_i$. Now $UF(X)=X\otimes A$ and $F(X)=\one\oplus R$, so $X^2=\one\oplus U(R)$. Finally \[ \dim\Hom_{\D}(Q_iQ_i^*,A) =\dim\Hom_{\N}(b_i^2R^2,\one)=b_i^2. \] The unit contributes one; Frobenius reciprocity identifies the remaining multiplicity with $N_{X,Q_i}^{Q_i}$. \end{proof} \begin{proposition}\label{B:squarefree} Even without the bimodule theorem below, the inputs imply $b_i^2\mid n+4$. Hence $b_i$ is odd, and \[ \rank\D=n+3-\sum_i(b_i^2-1)\equiv n+3\pmod8. \] In particular the asserted ranks hold whenever $n+4$ is squarefree. \end{proposition} \begin{proof} For a modular category, the column character $a\mapsto S_{ai}/S_{0i}$ of its integral fusion ring takes algebraic-integer values. Its formal codegree is the sum of the products of its values on $a$ and $a^*$, namely $1/S_{0i}^2$; this is an algebraic integer. This elementary form of the formal-codegree argument is also recorded in \cite{B:Ostrik}. Here \[ S_{0i}^2=\frac{b_i^2\delta^2}{n(2+\delta)^2} =\frac{b_i^2}{n+4}, \] using $\delta^2=n(1+\delta)$ and $(2+\delta)^2=(n+4)(1+\delta)$. Thus $(n+4)/b_i^2$ is a rational algebraic integer. The other conclusions follow from \eqref{B:squares}. \end{proof} For general $n$, the remaining task is to exclude $b_i>1$ when $n+4$ has a square divisor. The bimodule $A(X)$ supplies the required bound. \subsection{The Virasoro structure of the complement} Use the Fock spaces and blocks of Section~\ref{A:app:lat}. Write \[ h_j=nj(j+1),\qquad g_j=h_j+(2j+1)^2=(n+4)j(j+1)+1. \] Thus $M_j=M(c,h_j)$, $L_j=L(c,h_j)$, and $K_j=L(c,g_j)$, with the Verma and Fock structures in \eqref{A:lat:Verma}--\eqref{A:lat:Fock}. The charge filtration and its exact block projections are those of Lemmas~\ref{A:blocks:union} and \ref{A:lat:blocklemma}. Let $v_j=e^{(j+1)\beta}$ generate the embedded $M_j\subset B_j$, and let $W_j$ span the bottom of the unique $L_j\subset B_j$. The orthogonal decomposition $\mathcal U=V\oplus X$ gives \begin{equation}\label{B:Xblocks} X=\bigoplus_{j\geq0}X_j,\qquad 0\longrightarrow M_j\longrightarrow X_j\longrightarrow K_j\longrightarrow0. \end{equation} Indeed the projection of the embedded $M_j$ into $X_j$ is injective, since $M_j$ has no $L_j$ submodule. Lemma~\ref{A:lat:essential} identifies the socle and, by self-duality, the head of $X_j$ with $K_j$. Choose a bottom vector $x_j$ of $X_j$, a highest vector $\phi_j$ of its socle, and a homogeneous lift $u_j$ of the highest vector of its head. Then \[ \wt x_j=h_j,\qquad \wt\phi_j=\wt u_j=g_j,\qquad (u_j,\phi_j)\ne0. \] Every vector in $X_j$ is a sum of negative Virasoro descendants of $x_j$ and $u_j$. Since $n$ is odd, every $h_j$ is even and every $g_j$ is odd. This separation will make the reduction triangular. \subsection{Transition maps between Virasoro blocks} \begin{lemma}\label{B:couplings} The vectors can be chosen so that: \begin{enumerate}[label=(\alph*),leftmargin=7mm] \item $x_j$ has nonzero lowest-charge component proportional to $e^{-j\beta}$; the projection of $W_{-2nj-1}x_j$ to the bottom of $X_{j+1}$ is nonzero. \item For $j\geq1$, projection of $Y(W,z)\phi_j$ to $X_{j-1}$ has a nonzero coefficient proportional to $\phi_{j-1}$. \item The coefficient of $u_{j+1}$ in the head of the $X_{j+1}$ component of $W_{2n-1-(g_{j+1}-g_j)}u_j$ is nonzero. \end{enumerate} \end{lemma} \begin{proof} The charge of $L_1^\lambda$ never decreases. Invariance expresses the pairing of two lattice states through the weight-zero functional, which is supported in charges zero and one. Hence $(v_j,v_j)=0$, since both inputs have charge $j+1$ and all further charge shifts are nonnegative. The bottom form of $B_j$ is nondegenerate, while the $W_j$ line is nondegenerate. Its orthogonal projection of $v_j$ into $X_j$ therefore subtracts a \emph{nonzero} multiple of $W_j$: otherwise a nonzero vector in the one-dimensional nondegenerate orthogonal complement would be isotropic. This proves the assertion about $x_j$. Normalize $W=W_1$ to have lowest component $e^{-\beta}$. The lowest-charge contribution to $Y(W,z)x_j$ is a nonzero scalar times \[ Y(e^{-\beta},z)e^{-j\beta} =\epsilon_j z^{2nj}\bigl(e^{-(j+1)\beta}+\text{higher oscillator terms}\bigr), \quad\epsilon_j\ne0. \] Block projection to $B_{j+1}$ preserves this coefficient: on the charge $-(j+1)$ associated graded it is the identity. The action of $V$ preserves $X$. At weight $h_{j+1}$ its projection is thus a nonzero multiple of $x_{j+1}$. This proves (a). The lowest-charge component $\phi_j^0$ of $\phi_j$ is the nonzero singular vector in $\mathsf F_{j+1}=M_j$. The lowest-charge part of $Y(W,z)\phi_j$ is \[ Y(e^{-\beta},z)\phi_j^0\in\mathsf F_j\{z\}. \] It is nonzero: the lattice exponential is a charge shift composed with invertible creation and annihilation exponentials. Its projection to $L_{j-1}$ is zero. A nonzero projected Virasoro intertwiner would, by contragredient symmetry, give an intertwiner of type $\binom{K_j}{L_1\ L_{j-1}}$. Its image would be $C_1$-cofinite by \cite[Theorem 4.1.7]{A:CJORY}, contrary to irreducibility and non-$C_1$-cofiniteness of $K_j$. Thus the nonzero series takes values in the submodule $K_{j-1}$ of $\mathsf F_j$. Its leading coefficient is a highest vector: the Virasoro commutator formula with the primary $e^{-\beta}$ shows that every positive $L_r$ annihilates the first nonzero coefficient on the highest input $\phi_j^0$. Since $K_{j-1}$ is an irreducible Verma module, that coefficient is at its bottom $g_{j-1}$, not at a higher singular weight. Exact block projection preserves its charge-$j$ component. The filtered part of $B_{j-1}$ of charges at least $j$ is precisely its embedded Verma module; the next filtered part is zero. Hence the projected coefficient is a nonzero multiple of the socle vector $\phi_{j-1}$, proving (b). For an even-weight primary, invariance gives $(W_r a,b)=(a,W_{4n-2-r}b)$. Apply this to (b), pairing with $u_j$ and $\phi_{j+1}$. The adjoint mode has weight increase $g_{j+1}-g_j$, and the pairing is nonzero. Since pairing with the socle detects the head quotient, this is precisely (c). \end{proof} \subsection{The two-generator bound}\label{B:sec:twogen} For a $V$-module $M$, let $A(M)=M/O_V(M)$ be the Frenkel--Zhu bimodule. Put $x_l=[T]*-$ and $x_r=-*[T]$. For homogeneous $v$ the standard relations are \begin{equation}\label{B:ZhuVir} [L_{-r}v]=(-1)^r(rx_r-x_l+\wt v)[v]\quad(r\geq1). \end{equation} Also the higher circle relations reduce $[W_{-s}v]$, $s\geq2$, to classes of $W_rv$ with $r\geq-1$, and \begin{equation}\label{B:ZhuW} [W]*[v]=[W_{-1}v]+\sum_{a=1}^{2n}\binom{2n}{a}[W_{a-1}v]. \end{equation} These follow directly from the defining residue relations; they hold without assuming $v$ is Virasoro highest. \begin{theorem}\label{B:twogen} The bimodule $A(X)$ is generated over $A(V)$ on the left and right by the two classes $[x_0]$ and $[u_0]$, of weights zero and one. Consequently \begin{equation}\label{B:fusionbound} N_{X,S}^{T}\leq2 \end{equation} for any simple $V$-modules $S,T$. \end{theorem} \begin{proof} By \eqref{B:Xblocks} and \eqref{B:ZhuVir}, it is enough to reduce $[x_j],[u_j]$. Induct on their conformal weights. The two bottom generators are $x_0,u_0$. Their weights are distinct from all subsequent generator weights. For $x_{j+1}$ use Lemma~\ref{B:couplings}(a). The target weight increase from $x_j$ is $2n(j+1)$. For $j=0$, the mode is $W_{-1}$, which is handled by \eqref{B:ZhuW}. For $j\geq1$ the mode is at most $-2$, and a higher circle relation replaces it by modes of resulting weight at most $h_j+2n2n$, so a higher circle relation reduces the relevant mode to states of smaller weight. After removing the nonzero head coefficient, the remaining part of its block lies in the Verma module generated by $x_{j+1}$, whose weight is strictly smaller than $g_{j+1}$. Other blocks have smaller generator weights as well, because no $h_l$ equals $g_{j+1}$ and the $g_l$ are strictly increasing. Thus induction also solves for $[u_{j+1}]$. This proves the generation assertion. Strong generation by $T,W$ implies that $A(V)$ is generated by $[T],[W]$. Since $[T]$ is central it is commutative. Its simple modules, hence all simple $V$-module top spaces, are one-dimensional. The Frenkel--Zhu--Li injection \cite[Theorem 2.12]{A:AA} bounds an intertwining-operator space by the contraction of $A(X)$ with the two top-space characters. That contraction is generated by the images of the two displayed classes and has dimension at most two. This proves \eqref{B:fusionbound}. \end{proof} \begin{corollary}\label{B:rankthm} Every $b_i$ in Lemma~\ref{B:multiplicities} is one. There are exactly $M=(n+3)/2$ new simples $Q_i$, and the two ranks in Theorem~\ref{A:cat:main} hold. \end{corollary} \begin{proof} Apply \eqref{B:fusionbound} with $S=T=Q_i$. Equation \eqref{B:XX} gives $b_i^2-1\leq2$, so the positive integer $b_i$ is one. Now use \eqref{B:squares} and the semion factorization. Equation \eqref{B:XX} becomes $X^2=\bigoplus_{S\in\Irr(\D)}S$. \end{proof} In particular the missing objects have an intrinsic construction: split the complementary summand \[ Q=X^2\ominus\left(\one\oplus X\oplus\bigoplus_x Z_x\right). \] Its endomorphism algebra is $\C^M$, and its primitive idempotents give exactly the $Q_i$, each once. Their semion translates give the other $M$ missing $V$-modules. Every simple is self-contragredient. On the contragredient top, $o(T)$ has the same eigenvalue, and so does $o(W)$ because $W$ is primary of even weight. The two Zhu characters coincide; Zhu's bijection gives the isomorphism. Rationality makes $A(V)$ semisimple, so its commutativity and the rank theorem give $A(V)\cong\C^{2n+6}$ as abstract algebras. \section{Modular data}\label{sec:modular} Order the simples of $\D_n$ as $\one,X,Z_x,Q_y$, with $1\leq x\leq(n-1)/2$ and $1\leq y\leq(m-1)/2$. The next formulas state that, for $u=n^{-1/2}$ and $v=m^{-1/2}$, \begin{equation}\label{B:Sfull} S=\begin{pmatrix} (u+v)/2&(u-v)/2&u\mathbf1^{\!t}&-v\mathbf1^{\!t}\\ (u-v)/2&(u+v)/2&u\mathbf1^{\!t}&v\mathbf1^{\!t}\\ u\mathbf1&u\mathbf1&C_n&0\\ -v\mathbf1&v\mathbf1&0&C_m \end{pmatrix},\qquad u=n^{-1/2},\quad v=m^{-1/2}, \end{equation} \begin{equation}\label{B:Tfull} (C_r)_{ab}=\frac{2}{\sqrt r}\cos\frac{4\pi ab}{r},\qquad \Theta=\operatorname{diag}\left(1,1; e^{2\pi i x^2/n};e^{2\pi i y^2/m}\right). \end{equation} The matrix $\Theta$ records the ribbon twists. Use the positive square root of $\dim\D_n$ in the normalization of $S$. The end of this section recovers the character normalization and compares the data with the formulas of~\cite{A:EG,B:EG11}. \subsection{The known blocks} Normalize categorical $S$ by the positive root $\sqrt{\dim\D}=\sqrt n(2+\delta)$. Let $B$ be the restriction matrix from the local $A$-modules. Its only nonzero entries are \[ B_{\one,0}=B_{X,0}=1,\qquad B_{Z_x,x}=B_{Z_x,-x}=1. \] Kirillov--Ostrik's local-module modular-invariance theorem, which does not assume unitarity, gives \begin{equation}\label{B:branchS} S_{\D}B=B S_{\Pcat_n},\qquad (S_{\Pcat_n})_{xy}=n^{-1/2}e^{4\pi ixy/n} \end{equation} \cite[Theorems 4.1 and 4.5]{A:KO}. Twists on restriction agree. Since $d(Z_x)=2+\delta$ and $d(Q_i)=\delta$, the vacuum row is \[ S_{00}=(u+v)/2,\quad S_{0X}=(u-v)/2,\quad S_{0Z_x}=u,\quad S_{0Q_i}=-v. \] Taking each column of \eqref{B:branchS}, and using symmetry, now gives exactly \eqref{B:Sfull} except for its last block, temporarily denoted $K=(K_{ij})$. The known twists are $\theta_X=1$ and $\theta_{Z_x}=e^{2\pi ix^2/n}$. All fusion matrices are real symmetric because all simples are self-dual. Their eigenvalues $S_{ai}/S_{0i}$ are real; hence $S$ is real. Thus $S^2=1$, and its remaining blocks satisfy \begin{equation}\label{B:Krelations} K\mathbf1=-\frac1{\sqrt m}\mathbf1,\qquad K^2=I_M-\frac2m\mathbf1\mathbf1^t. \end{equation} The normalized Gauss sum of $\D$ equals that of its local pointed category \cite[Theorem 4.5]{A:KO}: \[ \gamma=\frac1{\sqrt n}\sum_{x\bmod n}e^{2\pi ix^2/n} =\begin{cases}1&n\equiv1\pmod4,\\ i&n\equiv3\pmod4.\end{cases} \] If $w_i=\theta_{Q_i}$, the vacuum entry of the Gauss-sum identity gives \begin{equation}\label{B:gaussmissing} 1+2\sum_{i=1}^{M}w_i=\gamma\sqrt m. \end{equation} For example, expand $\sum_a S_{0a}^2\theta_a=\gamma S_{00}$ and use the ordinary quadratic Gauss sum for the $Z$ sector. This proves \eqref{B:gaussmissing} without choosing a proposed twist for any $Q_i$. \subsection{The integral fusion algebra}\label{B:sec:metric} We recover the unknown block by adapting the characteristic-two method of \cite[Section~5.2]{B:GSY}; this also treats square $m$. Write $N_{ij}^k=N_{Q_i,Q_j}^{Q_k}$. Adjunction gives \begin{equation}\label{B:QQ} Q_iQ_j=\delta_{ij}\one+(1-\delta_{ij})X+\sum_x Z_x+\sum_k N_{ij}^kQ_k, \qquad \sum_kN_{ij}^k=M-2+\delta_{ij}. \end{equation} Indeed the coefficient of $Z_x=U(x)$ is the multiplicity of $x$ in $R^2$, the sum of the unit and $X$ coefficients is one, and the last equality follows by applying $F$ and comparing $R$ coefficients. Similarly \[ XQ_i=X+\sum_x Z_x+\sum_{j\ne i}Q_j. \] In the complexified fusion ring define $c_i=\one+X-Q_i$. Direct substitution using $X^2=\sum_a a$ gives \begin{equation}\label{B:cc} c_ic_j=2\delta_{ij}\one+\sum_k a_{ij}^kc_k,\qquad a_{ij}^k=1-N_{ij}^k\leq1,\qquad \sum_k a_{ij}^k=2-\delta_{ij}. \end{equation} The elements $\one,c_1,\ldots,c_M$ are linearly independent in the fusion ring. Evaluate at the fusion-ring character belonging to $Q_j$. The evaluations are \[ x_i(j)=\frac{S_{0j}+S_{Xj}-S_{Q_i j}}{S_{0j}}=\sqrt m K_{ij}. \] They form a basis of functions on the $M$ labels, since $K$ is invertible. Equations \eqref{B:Krelations} imply \begin{equation}\label{B:tracepair} \sum_i x_i=-1,\qquad \operatorname{Tr}(x_ix_j)=m\delta_{ij}-2. \end{equation} Here the trace is the sum over all $M$ evaluations. The lattice $\mathcal O=\sum_i\Z x_i$ is a unital integral algebra by \eqref{B:cc} and $1=-\sum_i x_i$. All its evaluations are algebraic integers, and its trace-pairing determinant is $m^{M-1}$, odd. Modular Galois symmetry \cite[Theorem 10.1]{B:ENO} permutes the $Q_i$ among themselves. Their vacuum entry has Galois conjugates $\pm1/\sqrt m$. The other vacuum entries have different absolute values, since \[ 0<\frac{u-v}{2}0. \] Since $d(Y)=-d(X)<0$, this forces the negative normalization of $S^V_{V,V}$ and hence the minus sign in \eqref{B:VOAmatrices}. This sign distinction matters for this nonunitary ribbon structure. \section{Characters and Zhu's algebra}\label{B:sec:characters} The complete module list is now known. The character determination uses the space of holomorphic modular forms of weight $1/2$ for the two Weil representations appearing in the modular data. This also treats composite parameters, where a single character need not generate the full modular representation. \subsection{Splitting the character vector into two Weil representations} For $r=n,m$ put \[ \vartheta_{a,r}=\eta\Lambda_{a,r} =\sum_{t\in\Z}q^{(2rt+a)^2/(4r)},\qquad 0\leq a\leq r. \] Let $\rho_r^+$ be the even part of the Weil representation of $(\Z/2r,a\mapsto a^2/(4r))$. Use component coordinates indexed by sign classes $0,1,\ldots,r$, so its theta transformation is \begin{align} \vartheta_{a,r}(-1/\tau)&=\frac{\sqrt{-i\tau}}{\sqrt{2r}} \left(\vartheta_{0,r}+(-1)^a\vartheta_{r,r} +2\sum_{b=1}^{r-1}\cos(\pi ab/r)\vartheta_{b,r}\right),\label{B:WeilS}\\ \vartheta_{a,r}(\tau+1)&=e^{2\pi ia^2/(4r)}\vartheta_{a,r}(\tau). \label{B:WeilT} \end{align} The factor $\sqrt{-i\tau}$ includes the usual $e^{-\pi i/4}$ in the metaplectic $S$ matrix. Temporarily label the $Z_x,Q_y$ and their $J$-translates by $A_a,B_b$ using \[ A_{n-2x}=Z_x,\quad A_{2x}=JZ_x,\qquad B_{m-2y}=Q_y,\quad B_{2y}=JQ_y. \] All $1\leq a0}$, so its fields have the integral grading required for a VOA. These checks make $Q$ an even simple-current unit. Carnahan's extension theorem \cite[Theorem~3.2.12]{E:Carnahan} now applies: its base is a simple $C_2$-cofinite M\"obius vertex algebra with finite weight spaces and a nondegenerate invariant form, and is rational. It gives a VOA extension on $B\oplus Q$ that is regular and $C_2$-cofinite. In particular, the theorem allows the present $B_0\neq\C\one$. More explicitly, choose the nonzero intertwining operator $\mathcal I$ of type $\binom{B}{Q\ Q}$ supplied by this construction. The four components of the state-field map are \begin{equation}\label{E:extensionfields} \begin{aligned} Y(b,z)b'&=Y_B(b,z)b',& Y(b,z)q&=Y_Q(b,z)q,\\ Y(q,z)b&=e^{zL(-1)}Y_Q(b,-z)q,& Y(q,z)q'&=\mathcal I(q,z)q', \end{aligned} \end{equation} for $b,b'\in B$ and $q,q'\in Q$. The vacuum and conformal vector lie in $B$. Equation~\eqref{E:Qcoherence} is precisely the compatibility of the last component with associativity and skew-symmetry; the extension theorem supplies the Jacobi identity for these fields. \emph{Simplicity and uniqueness.} As a $B$-module, $B\oplus Q$ is a multiplicity-free sum of two inequivalent simples. Any VOA ideal is a $B$-submodule. If it contains $B$, it contains the vacuum and hence the whole extension. If it contains $Q$, the nonzero $Q\times Q\to B$ product forces it to contain $B$. Thus the extension is simple. For any other simple extension on the same $B$-module, the first three components of \eqref{E:extensionfields} are fixed by the $B$-action and skew-symmetry. The last is $a\mathcal I$ for a scalar $a$, and $a\neq0$ since otherwise $Q$ would be an ideal. Choose $s\in\C^\times$ with $s^2=a$. The map $\id_B\oplus s\id_Q$ is an isomorphism from that extension to the one with last component $\mathcal I$: on $Q\times Q$ the required equality is exactly $a=s^2$. This proves uniqueness with $B$ fixed. \emph{Grading and central charge.} The $B$ summand has nonnegative integral weights, while $Q$ starts at the positive integer $h_Q\geq6$ and has integral weight increments. The conformal vector is that of $B$, giving \[ c(\mathscr{V}_N)=c(V)+c(\mathcal W)=(1-6n)-11=-6n-10. \] \end{proof} \begin{proposition}\label{E:selfdual} The extension $\mathscr{V}_N$ is self-contragredient. \end{proposition} \begin{proof} Since $h_Q\geq6$, its degree-zero and degree-one spaces coincide with those of $B$, with the same $L(1)$ action. Consequently \[ (\mathscr{V}_N)_0/L(1)(\mathscr{V}_N)_1=B_0/L(1)B_1. \] The space of invariant bilinear forms on $B$ is $\Hom_B(B,B')$, where $B'$ is its restricted contragredient. Since $B$ is simple and self-contragredient, this space has dimension one. Li's invariant-form theorem \cite{A:Li} identifies it with the dual of $B_0/L(1)B_1$. Applying the same theorem to $\mathscr{V}_N$ therefore gives a one-dimensional space of invariant forms on the extension. Choose a nonzero form in this space. Its radical is an ideal; simplicity from Proposition~\ref{E:existence} forces that radical to be zero. Invariance makes distinct $L(0)$-eigenspaces orthogonal, and each eigenspace is finite-dimensional. The nondegenerate form thus gives an isomorphism $\mathscr{V}_N\cong\mathscr{V}_N'$ of modules. No normalization of the vacuum's self-pairing is needed. \end{proof} \subsection{Modularity without CFT type}\label{E:nonCFTsection} \begin{proposition}\label{E:nonCFTmodular} The category of grading-restricted $\mathscr{V}_N$-modules is a semisimple modular tensor category, despite $\dim (\mathscr{V}_N)_0=56$. \end{proposition} \begin{proof} Proposition~\ref{E:existence} gives a simple, regular, $C_2$-cofinite, nonnegatively $\Z$-graded VOA $\mathscr{V}_N$, and Proposition~\ref{E:selfdual} proves it is self-contragredient. In particular, regularity implies rationality and hence semisimplicity of Zhu's algebra $A(\mathscr{V}_N)$. McRae's Main Theorem~3 \cite{A:McR} applies to precisely an $\mathbb N$-graded, simple, self-contragredient, $C_2$-cofinite VOA with semisimple Zhu algebra. It therefore proves that the full category of grading-restricted generalized $\mathscr{V}_N$-modules is a semisimple modular tensor category. No hypothesis that $(\mathscr{V}_N)_0=\C\one$ is used. This argument establishes modularity independently of the subsequent identification of the category with $\D_n$; in particular, that identification is not used to deduce the rigidity required in McRae's modularity results. \end{proof} \subsection{The diagonal algebra and its local modules}\label{E:localmodules} Under \eqref{E:basecat}, the extension algebra is \begin{equation}\label{E:diagonal} \mathscr A=\one\oplus Q =\one_{\D_n}\bo\bigl((\one\bo\one)\oplus(J\bo M)\bigr). \end{equation} The factor in parentheses is the diagonal algebra in $\Scat^-\bo(\Scat^-)^{\mathrm{rev}}$. Equations~\eqref{E:Qdata}--\eqref{E:Qcoherence} identify the symmetric subcategory generated by $Q$ with $\Rep(\Z/2)$. The multiplication on $\one\oplus Q$ is its regular commutative algebra, so it is separable over $\C$. It is connected because \[ \Hom_{\Rep(B)}(B,\mathscr A)=\Hom_B(B,B)\cong\C; \] connectedness here concerns the multiplicity of the tensor unit, not the dimension of the weight-zero space. Moreover $\theta_{\mathscr A}=\id$ and $d(\mathscr A)=2$. Thus $\mathscr A$ is a connected \'etale algebra with trivial twist, and its module category is semisimple. We next calculate its local modules and their ribbon structure explicitly. \begin{proposition}\label{E:category} There is a ribbon equivalence \[ \Rep(\mathscr{V}_N)\simeq\D_n\simeq Z(\Fcat_N). \] For each simple $T\in\D_n$, the corresponding irreducible ordinary $\mathscr{V}_N$-module is \begin{equation}\label{E:ET} \mathsf M_T=(T\otimes \mathcal W)\oplus((J\bo_V T)\otimes M). \end{equation} These modules are pairwise nonisomorphic and exhaust the irreducible ordinary modules. \end{proposition} \begin{proof} \emph{The extension correspondence.} Put $\mathcal C=\Rep(B)$. Since $B$ is $\mathbb N$-graded and $C_2$-cofinite, its full category of grading-restricted generalized modules has the Huang--Lepowsky--Zhang vertex tensor structure \cite[Section~2.3]{A:McR}. Regularity makes this category semisimple. Proposition~2.17 of the same reference identifies the braided structure before and after the lattice conformal shift, as checked in Lemma~\ref{E:inverse}; thus it is the product structure in \eqref{E:basecat}. The extension $\mathscr{V}_N=B\oplus Q$ is an object of this category with the same conformal vector as $B$. The extension theorem of Creutzig--Kanade--McRae \cite[Theorem~3.65]{E:CKMext} consequently gives a braided equivalence \begin{equation}\label{E:localcorrespondence} \mathcal C_{\mathscr A}^{\mathrm{loc}}\simeq\Rep(\mathscr{V}_N), \end{equation} initially for those modules whose restrictions lie in $\mathcal C$. This accounts for the full grading-restricted category. Indeed, an ordinary $\mathscr{V}_N$-module restricts, by regularity of $B$, to a direct sum of simple ordinary $B$-modules. There are finitely many such types. Infinite multiplicity of any one type would give an infinite-dimensional $L(0)$-eigenspace, because the conformal vector has not changed. Its restriction is therefore a finite direct sum and belongs to $\mathcal C$. Regularity of $\mathscr{V}_N$ also makes every grading-restricted generalized module a finite direct sum of ordinary simples. Conversely a local $\mathscr A$-module in $\mathcal C$ acquires the extension action with the same grading, so is grading-restricted. This proves \eqref{E:localcorrespondence} for precisely the category in the proposition. \emph{Locality and the simple objects.} Write the simple objects of $\mathcal C$ as \[ S_{T,a,b}=(J^{\bo_V a}\bo_V T)\otimes M^{\bo_{\mathcal W}b}, \qquad T\in\Irr(\D_n),\quad a,b\in\{0,1\}, \] where the zeroth powers mean $V$ and $\mathcal W$. In either semion factor the monodromy of the nontrivial object with itself is $-1$. Since $T$ centralizes $J$, the monodromy with $Q$ is \begin{equation}\label{E:localitytest} c_{S_{T,a,b},Q}\,c_{Q,S_{T,a,b}} =(-1)^{a+b}\id_{Q\bo_B S_{T,a,b}}. \end{equation} Let $\operatorname{Ind}(S)=\mathscr A\bo_B S$ denote free induction. By \cite[Proposition~2.65]{E:CKMext}, it is local exactly when the monodromy of $S$ with $\mathscr A$ is trivial. Formula~\eqref{E:localitytest} says that this is exactly $a=b$. Tensoring with $Q$ sends $(T,a,b)$ to $(T,a+1,b+1)$, with the last two labels read modulo two. Hence the local labels occur in the free pairs \[ \{S_{T,0,0},S_{T,1,1}\},\qquad T\in\Irr(\D_n). \] There are no fixed labels, since the second lattice factor changes from $\mathcal W$ to $M$ or conversely. Induction--restriction adjunction gives, for every simple $S$, \begin{equation}\label{E:inductionend} \End_{\mathscr A}(\operatorname{Ind}(S)) \cong\Hom_{\mathcal C}(S,S\oplus(Q\bo_B S))\cong\C. \end{equation} The last equality uses the absence of fixed labels. Since the $\mathscr A$-module category is semisimple, this proves that each induced module is simple. Induction identifies the two labels in each pair: the isomorphism $\mathscr A\bo_B Q\cong\mathscr A$ follows from its regular action. For the representative $S_{T,0,0}=T\otimes\mathcal W$, the underlying $B$-module is exactly \[ \operatorname{Ind}(T\otimes\mathcal W) =(T\otimes\mathcal W)\oplus((J\bo_V T)\otimes M), \] as in \eqref{E:ET}. The same adjunction gives \begin{equation}\label{E:inductionhom} \Hom_{\mathscr A}(\mathsf M_T,\mathsf M_{T'}) \cong\Hom_{\D_n}(T,T'), \end{equation} because the cross term has lattice factors $\mathcal W$ and $M$, which are nonisomorphic. This proves pairwise nonisomorphism. To prove exhaustion, let $L$ be a simple local $\mathscr A$-module. Its action $Q\bo_B L\to L$ is invertible: acting twice is the nonzero multiplication $Q\bo_B Q\cong B$ followed by the unit action. Locality then forces the monodromy with $Q$ on its underlying object to be the identity. Any simple $B$-constituent $S$ of $L$ therefore has $a=b$. Adjunction gives a nonzero $\mathscr A$-module map $\operatorname{Ind}(S)\to L$. Both modules are simple by \eqref{E:inductionend}, so this map is an isomorphism. Thus every local simple is one of the displayed $\mathsf M_T$. \emph{The tensor product and braiding.} Define \begin{equation}\label{E:ribbonfunctor} \Phi:\D_n\longrightarrow\Rep(\mathscr{V}_N),\qquad \Phi(T)=\operatorname{Ind}(T\otimes\mathcal W). \end{equation} Every object $T\otimes\mathcal W$ centralizes $\mathscr A$. Induction on this centralizer is a braided tensor functor \cite[Theorems~2.67 and 3.68]{E:CKMext}. In particular its tensor and unit isomorphisms are \begin{equation}\label{E:inductiontensor} \Phi(T)\bo_{\mathscr{V}_N}\Phi(T')\cong\Phi(T\bo_V T'), \qquad \Phi(V)\cong\mathscr{V}_N. \end{equation} These isomorphisms arise from multiplication of the two $\mathscr A$ factors in the relative tensor product. The centralizer condition makes them compatible with the braiding, and associativity of $\mathscr A$ gives their tensor coherence. Equation~\eqref{E:inductionhom}, extended over direct sums, proves full faithfulness; the preceding exhaustion argument proves essential surjectivity. Thus $\Phi$ is a braided tensor equivalence, rather than merely a matching of fusion rules or modular matrices. \emph{The ribbon structure and the center.} The twist on a VOA module is $e^{2\pi i L(0)}$. On the first summand of $\mathsf M_T$ it is $\theta_T$, since $\theta_{\mathcal W}=1$. On the second, centralization of $J$ by $T$ gives \begin{equation}\label{E:twistpreserved} \theta_{J\bo_V T}\theta_M =\theta_J\theta_T\theta_M=(-i)i\theta_T=\theta_T. \end{equation} Thus $\Phi$ preserves the twist on each simple and hence on every object. The two categories are rigid, by the results for $V$ and Proposition~\ref{E:nonCFTmodular}, so this proves that $\Phi$ is a ribbon equivalence. Finally Section~\ref{A:cat:condensation} constructs the spherical category $\Fcat_N$ and proves the ribbon equivalence $\D_n\simeq Z(\Fcat_N)$ in \eqref{A:cat:centerF}. Composing it with $\Phi^{-1}$ proves the asserted realization of the center, with its inherited spherical structure and associators. \end{proof} Together with Proposition~\ref{E:nonCFTmodular}, this proves the existence and categorical parts of Theorem~\ref{E:main} without applying a CFT-type modularity theorem to $\mathscr{V}_N$. The fusion coefficients of the $\mathsf M_T$ are exactly those of $\D_n$, and the $S$ matrix is the categorical matrix of Section~\ref{sec:modular}, with its stated normalization. The categorical twists should be distinguished from the trace $T$ matrix, which also includes the factor $e^{-2\pi i c(\mathscr{V}_N)/24}$. \section{A rank-eight lattice parent}\label{E:lattice} This section uses the concrete lattice embedding and the half-lattice decomposition of Section~\ref{A:app:lat}. It gives actual vertex operators for \eqref{E:def}, in addition to the abstract simple-current construction. \subsection{The lattice and its discriminant form} Choose a root $\alpha\in E_8$, with $\alpha^2=2$, and identify $K=E_8\cap\alpha^\perp$ with $E_7$. Choose the sign of $u$ so that \[ E_8=(\Z\alpha\oplus K)+\Z(\alpha/2+u). \] Set $\beta=N\alpha$ and \begin{equation}\label{E:LN} L_N=(\Z\beta\oplus K)+\Z(\beta/2+u),\qquad h_N=\beta/2+u. \end{equation} The new coset has order two over $\Z\beta\oplus K$, and \[ h_N^2=\frac{N^2+3}{2}\in2\Z. \] Thus $L_N$ is even and positive definite of rank eight. Its determinant is $(2N^2\cdot2)/2^2=N^2$. \begin{lemma}\label{E:discriminant} There are equalities and an isometry of finite quadratic modules \begin{align} L_N&=\{x\in E_8:(x,\alpha)\equiv0\pmod N\},\label{E:kernel}\\ (L_N^*/L_N,q)&\cong(\Z/N^2,\ x\longmapsto x^2/N^2\pmod\Z).\label{E:discriminantform} \end{align} In particular $[E_8:L_N]=N$. \end{lemma} \begin{proof} Every generator in \eqref{E:LN} has pairing with $\alpha$ divisible by $N$, and $h_N\in E_8$ because $N$ is odd. The homomorphism $E_8\to\Z/N$ given by $x\mapsto(x,\alpha)$ is surjective: roots of $E_8$ have pairing one with $\alpha$. Its kernel has index $N$, which agrees with the determinant of $L_N$, proving \eqref{E:kernel}. The vector $\alpha/N$ lies in $L_N^*$ by \eqref{E:kernel}. Moreover $L_N\cap\R\alpha=N\Z\alpha$: a vector in the odd glue coset has a nontrivial $K^*/K$ component, and cannot lie on the $\alpha$ line. Therefore $\alpha/N$ has exact order $N^2$ modulo $L_N$. Since the discriminant group has order $N^2$, it is generated by this element. Its quadratic value is $(\alpha/N)^2/2=1/N^2$, proving the formula. \end{proof} With the standard conformal vector, this is the concrete inner cyclic orbifold \begin{equation}\label{E:orbifold} V_{L_N}=V_{E_8}^{\langle\exp(2\pi i\alpha(0)/N)\rangle}. \end{equation} Indeed the automorphism fixes oscillators and acts on $e^x$ by $\exp(2\pi i(x,\alpha)/N)$. Its fixed momenta are exactly \eqref{E:kernel}. The subsequent conformal deformation is essential for the embedding of the diagonal family. \subsection{The conformal vector and the subalgebra} Fix $\lambda\ne0$ and put \begin{equation}\label{E:ambientomega} \omega_N^{\mathrm{amb}} =\omega_{L_N}+h_N(-2)\one+\lambda D^2e^\beta. \end{equation} For the first two terms, $e^\beta$ is primary of weight zero, because $\beta^2/2=(h_N,\beta)=n$. Its self-OPE is regular. If $L^{(h_N)}_r$ denotes the shifted lattice Virasoro operator, the additional term gives \[ L^{\mathrm{amb}}_r=L^{(h_N)}_r+\lambda r(r+1)(e^\beta)_{r-1}. \] For a weight-zero primary, $[L_r,(e^\beta)_s]=(-r-s-1)(e^\beta)_{r+s}$; the modes of $e^\beta$ commute with one another. Substitution proves the Virasoro relations, since the cross-term coefficient is \[ \lambda(r+s)\bigl(r(r+1)-s(s+1)\bigr) =\lambda(r-s)(r+s)(r+s+1). \] There is no added central term. Both $L(0)$ and $L(-1)$ are unchanged by $\lambda D^2e^\beta$. The resulting regular VOA $\mathcal P_N$ has \[ c(\mathcal P_N)=8-12h_N^2=-6N^2-10. \] Regularity follows again from the unchanged lattice fields and the ordinary shifted grading of every lattice module. The two lattice cosets in \eqref{E:LN} give \begin{equation}\label{E:Pbranch} \mathcal P_N=(\mathcal U_n\otimes \mathcal W) \oplus(\mathcal M_n\otimes M), \end{equation} where $\mathcal U_n=V\oplus X$ and $\mathcal M_n=J\oplus Y$ are the deformed rank-one lattice VOA and its half-lattice module. The $V\otimes\mathcal W$ action preserves $J\otimes M$. For the remaining product, $J\bo_V J=V$ and $M\bo_{\mathcal W}M=\mathcal W$ force the image of $(J\otimes M)\times(J\otimes M)$ into $V\otimes\mathcal W$. This product is nonzero: its first factor is the nonzero lattice intertwiner on $J\times J$ proved in Section~\ref{A:app:lat}, and its second factor is the lattice intertwiner $M\times M\to\mathcal W$. Thus all products in \[ (V\otimes \mathcal W)\oplus(J\otimes M) \] are closed under the actual lattice fields, and the subspace contains the vacuum and conformal vector \eqref{E:ambientomega}. It is therefore a conformal subalgebra of $\mathcal P_N$. The two-summand ideal argument in Proposition~\ref{E:existence} proves that it is simple, and the uniqueness assertion of that proposition identifies its vertex operators with those of $\mathscr{V}_N$, fixing $V\otimes\mathcal W$. This supplies a concrete lattice realization of the extension fields \eqref{E:extensionfields}. The complementary terms of \eqref{E:Pbranch} give the precise branching \begin{equation}\label{E:parentbranch} \mathcal P_N\downarrow \mathscr{V}_N=\mathscr{V}_N\oplus\mathsf M_X, \qquad \mathsf M_X=(X\otimes \mathcal W)\oplus(Y\otimes M). \end{equation} The inclusion $L_N\subset E_8$ gives a further conformal extension with the same vector \eqref{E:ambientomega}. Its underlying vertex algebra is $V_{E_8}$; with this conformal vector it is regular and has only one irreducible ordinary module. This use of holomorphicity permits negative conformal weights. Both ambient lattice algebras have minimum weight \[ -\frac{h_N^2}{2}=-\frac{N^2+3}{4}, \] attained at $e^{h_N}$. In contrast $\mathscr{V}_N$ itself is nonnegatively graded. The discriminant group in \eqref{E:discriminantform} is cyclic of order $N^2$; it is different from the split group $(\Z/N)^2$ occurring in an untwisted pointed parent. \section{Characters and examples}\label{E:characters} For a lattice coset write \[ \Theta_{K+a}(\tau)=\sum_{p\in K+a}q^{p^2/2},\qquad \theta_{r,s}(\tau)=\sum_{\ell\in\Z}q^{(2s\ell+r)^2/(4s)}, \qquad q=e^{2\pi i\tau}. \] The character convention is $\ch F=\operatorname{tr}_F q^{L(0)-c/24}$. Completing the square in \eqref{E:shiftweight} gives \begin{equation}\label{E:auxchars} \ch \mathcal W=\frac{\Theta_{K+u}}{\eta^7},\qquad \ch M=\frac{\Theta_K}{\eta^7}. \end{equation} Thus the conformal shift exchanges the two standard $E_7$ characters. \subsection{All irreducible characters}\label{E:allcharacters} By \eqref{E:ET}, the character attached to $T\in\Irr(\D_n)$ is \begin{equation}\label{E:allchars} \ch\mathsf M_T =\frac{\chi_T\Theta_{K+u}+\chi_{JT}\Theta_K}{\eta^7}. \end{equation} These are traces on constructed modules. Theorem~\ref{B:characterthm} makes \eqref{E:allchars} completely explicit. With $m=n+4$, the vacuum and the distinguished non-vacuum character are \begin{align} \ch \mathscr{V}_N&= \frac{(\theta_{n,n}-\theta_{m,m})\Theta_{K+u} +(\theta_{0,n}-\theta_{0,m})\Theta_K}{2\eta^8},\label{E:vacchar}\\ \ch\mathsf M_X&= \frac{(\theta_{n,n}+\theta_{m,m})\Theta_{K+u} +(\theta_{0,n}+\theta_{0,m})\Theta_K}{2\eta^8}.\label{E:Xchar} \end{align} For odd $a$ with $1\leq a0$ eventually. No existence assertion for such a $P$ is made here. Nor does this sufficient auxiliary problem exclude other constructions of CFT-type realizations. \subsection{The twisted target} The invariant \[ \prod_{b=0}^{N-1}\Omega_N(1,b,1)=e^{2\pi i/N} \] shows that the pointed associator is nontrivial for every parameter under consideration. The cancellation in Section~\ref{E:extension} preserves $\Fcat_N$ and its center; it does not remove this cohomology class or change $d(\rho)$ to its positive Frobenius--Perron value. There is also a distinction directly in the center. Section~\ref{sec:modular} gives an object of twist $e^{2\pi i/N^2}$, of exact order $N^2$. For the canonical untwisted Haagerup--Izumi data the twist orders divide $N$ or $N^2+4$ \cite[Section~3.2]{B:EG11}; none has order $N^2$, because $\gcd(N,N^2+4)=1$ for odd $N$. A cyclotomic Galois automorphism preserves the exact order of a root of unity, so Galois conjugation cannot erase this distinction. \subsection{The nonunitary ribbon structure} The category $\D_n$ has negative-dimensional objects $Q_y$, with \[ d(Q_y)=\frac{n-\sqrt{n(n+4)}}2<0. \] Hence its specified ribbon structure cannot be the representation category of a unitary VOA. This is separate from the elementary fact that the constructed $\mathscr{V}_N$ has negative central charge. The signs cannot be changed by a different pivotal structure on the same underlying fusion category. Corollary~\ref{B:rankthm} gives \[ X^2=\bigoplus_{T\in\Irr(\D_n)}T, \] including both $\one$ and $X$. If a tensor automorphism of the identity has scalar $\lambda_T$ on $T$, then every displayed summand satisfies $\lambda_T=\lambda_X^2$, while the $X$ summand gives $\lambda_X=\lambda_X^2$. Thus $\lambda_X=1$ and every $\lambda_T=1$. Pivotal structures, when they exist, differ by such automorphisms; the existing one is unique. Finally the Haagerup--Izumi fusion rules themselves allow no nontrivial group grading of $\Fcat_N$. The summand $\rho$ of $\rho^2$ forces $\deg\rho=e$; the other summands force $\deg\alpha_a=e$. Consequently a cocycle twist of a grading of the entire category does not provide an automatic inverse to \eqref{E:HIinvariants}. A realization of the canonical untwisted unitary series remains a different reconstruction problem. \appendix \section{The level-nine singular-vector calculation}\label{app:null} This appendix gives the coefficient data used in Section~\ref{A:app:c2}. The singular vector is normalized by the coefficient of $L_{-1}^9W$; the recurrence follows by applying the Ward operators in PBW order. \subsection{The singular vector} Use PBW order $L_{-\lambda}=L_{-\lambda_1}\cdots L_{-\lambda_\ell}$ with $\lambda_1\geq\cdots\geq\lambda_\ell$. In \[ s_{3,3}=\sum_{\lambda\vdash9}c_\lambda(n)L_{-\lambda}W,\qquad c_\lambda(n)=\sum_{a=0}^6c_{\lambda,a}n^a, \] all coefficients are as follows. A repeated part is written as a power; for example $(2,1^7)$ has eight parts. \Needspace{8\baselineskip} {\small\setlength{\tabcolsep}{6pt} \begin{longtable}{@{}lrrrrrrr@{}} \toprule $\lambda$&$c_0$&$c_1$&$c_2$&$c_3$&$c_4$&$c_5$&$c_6$\\\midrule \endfirsthead \toprule $\lambda$&$c_0$&$c_1$&$c_2$&$c_3$&$c_4$&$c_5$&$c_6$\\\midrule \endhead $(9)$ & 0 & -2016 & 12544 & 0 & -1792 & -2048 & 0 \\ $(8,1)$ & 432 & -7936 & 2464 & -3136 & 2432 & -256 & 0 \\ $(7,2)$ & 0 & 1664 & -8192 & 4544 & 1280 & -256 & 0 \\ $(7,1^{2})$ & -384 & 1008 & 2832 & -3552 & 384 & 0 & 0 \\ $(6,3)$ & 0 & -3392 & 18432 & -2816 & -5632 & -1024 & 0 \\ $(6,2,1)$ & 64 & -2048 & -6272 & 3072 & 1536 & 0 & 0 \\ $(6,1^{3})$ & 480 & -1280 & 1920 & -640 & 0 & 0 & 0 \\ $(5,4)$ & 0 & 416 & -5312 & 2432 & 128 & -256 & 0 \\ $(5,3,1)$ & 808 & -7664 & 2824 & 3696 & 96 & -192 & 0 \\ $(5,2^{2})$ & 0 & 1536 & -1536 & 960 & 384 & 0 & 0 \\ $(5,2,1^{2})$ & -1328 & 4256 & -3600 & -960 & 192 & 0 & 0 \\ $(5,1^{4})$ & 288 & -440 & 472 & -48 & 0 & 0 & 0 \\ $(4^{2},1)$ & -624 & 7968 & -3648 & -192 & 384 & 0 & 0 \\ $(4,3,2)$ & 0 & -2560 & 11392 & -288 & -2304 & -384 & 0 \\ $(4,3,1^{2})$ & 312 & -624 & -3096 & 96 & 288 & 0 & 0 \\ $(4,2^{2},1)$ & 256 & -9088 & -1152 & 1728 & 384 & 0 & 0 \\ $(4,2,1^{3})$ & 864 & 3520 & -224 & -384 & 0 & 0 & 0 \\ $(4,1^{5})$ & -216 & -180 & 72 & 0 & 0 & 0 & 0 \\ $(3^{3})$ & 0 & 1152 & -5088 & -424 & 1904 & 672 & 64 \\ $(3^{2},2,1)$ & -256 & 3328 & 2208 & -2832 & -1536 & -192 & 0 \\ $(3^{2},1^{3})$ & -324 & 176 & 260 & 192 & 48 & 0 & 0 \\ $(3,2^{3})$ & 0 & 0 & -1024 & -256 & 0 & 0 & 0 \\ $(3,2^{2},1^{2})$ & 384 & -1344 & 1728 & 1248 & 192 & 0 & 0 \\ $(3,2,1^{4})$ & -144 & -240 & -336 & -96 & 0 & 0 & 0 \\ $(3,1^{6})$ & 36 & 6 & 12 & 0 & 0 & 0 & 0 \\ $(2^{4},1)$ & 0 & 1024 & 256 & 0 & 0 & 0 & 0 \\ $(2^{3},1^{3})$ & -256 & -640 & -384 & -64 & 0 & 0 & 0 \\ $(2^{2},1^{5})$ & 144 & 192 & 48 & 0 & 0 & 0 & 0 \\ $(2,1^{7})$ & -24 & -12 & 0 & 0 & 0 & 0 & 0 \\ $(1^{9})$ & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ \bottomrule \end{longtable}} The coefficient of $L_{-1}^9W$ is one. The verification uses only \[ [L_a,L_b]=(a-b)L_{a+b}+\frac{1-6n}{12}(a^3-a)\delta_{a+b,0},\qquad L_0W=2nW, \] and $L_mW=0$ for $m>0$. Negative-mode reordering uses $L_{-a}L_{-b}=L_{-b}L_{-a}+(b-a)L_{-(a+b)}$ for $a1$, and $m=n+4$. Let $W$ be the primary generator of weight $h$, and let $s_\lambda(h)=c_\lambda(h/2)$ be the thirty singular-vector coefficients in Appendix~\ref{app:null}. Write $\nu=W_{2h-1}W\in\C^\times$ and let $C\ne0$ be the coefficient of $W$ in $W_{h-1}W$. Their nonvanishing is supplied by the lattice construction. The notation $x=[T]$, $y=[W]$ will be used in either the $C_2$ algebra or Zhu's algebra, as specified. \subsection{Complete recursive coefficient definitions} For a partition $\lambda=(r,\mu)$ of a number at most nine and an integer $s\geq0$, define polynomials in an auxiliary variable $z$ by \begin{align*} F_{\varnothing,s}(a,h;z)&=1,\\ F_{(r,\mu),s}(a,h;z)&=\delta_{r,2}zF_{\mu,s}(a,h;z) +\bigl(2h+8-2a-s+(h-1)(r-1)\bigr)F_{\mu,s+r}(a,h;z),\\ Q_p(a,h)&=[z^p]\sum_{\lambda\vdash9}s_\lambda(h)F_{\lambda,0}(a,h;z). \end{align*} Only $0\leq p\leq4$ occur. For Zhu's algebra define \begin{align*} G_{\varnothing,s}(d,h,x;z)&=z^{9-s},\\ G_{(r,\mu),s}(d,h,x;z)&=(-1)^r\bigl((r-1)x+d-9+s+|\mu|\bigr)G_{\mu,s}(d,h,x;z)\\ &\quad+\bigl(2h+8-d-s+(h-1)(r-1)\bigr)G_{\mu,s+r}(d,h,x;z),\\ K_p(d,h,x)&=[z^p]\sum_{\lambda\vdash9}s_\lambda(h)G_{\lambda,0}(d,h,x;z). \end{align*} Only $0\leq p\leq9$ occur, $K_0(d,h,x)=Q_0(d/2,h)$, and $\deg_xK_p\leq\lfloor p/2\rfloor$. These fixed recursions involve the thirty partitions of nine. Commuting $W_a$ through the singular vector before passing to the relevant quotient gives \begin{align} \sum_{p=0}^4Q_p(a,h)x^pR_{a-p}&=0, &R_a&=[W_{2h-1-2a}W]\in V/C_2(V),\label{H:C2rec}\\ \sum_{p=0}^9K_p(d,h,x)Z_{d-p}&=0, &Z_d&=[W_{2h-1-d}W]\in A(V).\label{H:Zhurec} \end{align} Negative indices mean zero. The identities follow from \[ [W_a,L_{-r}]=\bigl(a+(h-1)(r-1)\bigr)W_{a-r},\qquad [L_{-r}v]=(-1)^r\bigl((r-1)x+\wt v\bigr)[v] \] in Zhu's algebra. In the $C_2$ quotient a leftmost $L_{-r}$ survives only for $r=2$. Thus the calculation does not assume that positive modes preserve $C_2(V)$. \subsection{The sharper uniform \texorpdfstring{$C_2$}{C2} bound} Here $R_a=0$ for $a<0$ or $a>h$, $R_0=\nu$, and $R_h=y^2$. The recursive coefficients give \[ Q_4(a,h)=-128h(h+8)(a-h-4). \] Define $b_0=1$, $b_j=0$ for $j<0$, and \begin{equation}\label{H:backward} b_j=-\frac{\sum_{p=0}^3Q_p(h+4-j,h)b_{j+p-4}}{128h(h+8)j} \quad(j\geq1). \end{equation} Put $B_j=2^j j!b_j/h^2$ for $j\geq1$. The first values are \begin{align*} B_1&=1,\\ B_2&=h^2-h+3,\\ B_3&=h^4-3h^3+3h^2+32h+57,\\ B_4&=h^6-6h^5-3h^4+200h^3-279h^2+6750h+3417. \end{align*} For $j\geq5$, \[ B_j=\sum_{q=1}^4A_q(h,j)B_{j-q},\qquad A_q(h,j)=-\frac{2^q(j-1)\cdots(j-q+1)Q_{4-q}(h+4-j,h)}{128h(h+8)}. \] For $h\equiv2\pmod4$, the following are lower bounds on $v_2(A_q)$; starred bounds are equalities: \[ \begin{array}{c|rrrr} j\pmod4&q=1&q=2&q=3&q=4\\\hline 0&1&1&2&2^*\\ 1&2&4&4&5\\ 2&2&2&4&4\\ 3&1&2&2^*&5 \end{array} \] They are finite polynomial congruences: substitute $h=4v+2$, $j=4r+s$ in the numerator of $A_q$. Its denominator is $2^9(2v+1)(2v+5)$. After division by $2^{9+e_{s,q}}$, all coefficients are integral; for $(s,q)=(0,4),(3,3)$ the numerator divided by $2^{11}$ is identically one modulo two. These statements are completely specified by the preceding recursion and the table of $s_\lambda$. Induction in blocks of four now gives \begin{equation}\label{H:valuation} v_2(B_j)\geq2\left\lfloor\frac{j-1}{4}\right\rfloor, \end{equation} with equality for $j\equiv0,3\pmod4$. The first four values are odd. In a block beginning at $4r+1$, the first two new values have valuation at least $2r$. For $j=4r+3$, the $q=3$ term has valuation exactly $2r$, and the other terms have larger valuation; for $j=4r+4$, the $q=4$ term is the unique term of valuation $2r$. This proves \eqref{H:valuation}, and in particular \[ v_2(B_{h+1})=n-1,\qquad b_{h+1}\ne0. \] \begin{theorem}[Sharper uniform finiteness bound]\label{H:C2bound} For every odd $n\geq3$, \[ x^{2n+3}=x^{n+3}y=0,\qquad \dim V_{n+2}/C_2(V_{n+2})\leq3n+6. \] \end{theorem} \begin{proof} Put $T_j=x^jR_{h-j}$ for $0\leq j\leq h+1$, and $T_j=0$ at negative indices. Multiply \eqref{H:C2rec} at $a=h+4-j$ by $x^{j-1}$. The resulting relation, compared with \eqref{H:backward}, proves inductively \[ x^3T_j=b_jx^3R_h. \] Since $T_{h+1}=0$ and $b_{h+1}\ne0$, one gets $x^3R_h=0$. At $j=h$ and $j=n$ this gives $\nu x^{h+3}=0$ and $x^{n+3}R_n=0$. The relation $R_n=\alpha x^n+Cy$ with $C\ne0$ implies both asserted vanishings. Also $y^2\in\C x^{2n}+\C x^ny$. The monomials \[ 1,x,\ldots,x^{2n+2},\qquad y,xy,\ldots,x^{n+2}y \] span the quotient and number $3n+6$. \end{proof} \subsection{A universal Zhu polynomial} Write $\mathcal B=A(V)$, a finite-dimensional commutative algebra generated by $x,y$. Set $Z_d$ as in \eqref{H:Zhurec}; then $Z_0=\nu\ne0$ and $Z_h=Cy+b(x)$ with $C\ne0$. There is a polynomial $P(w)\in\mathcal B[w]$ with \begin{equation}\label{H:P} \sum_{d\geq0}Z_dz^d=P\!\left(\frac{z^2}{1+z}\right),\qquad \deg P\leq h,\qquad P(0)=\nu\one. \end{equation} Indeed higher Zhu circle relations make $(1+z)^h\sum_dZ_dz^d$ a polynomial of degree at most $2h$. Skew symmetry and $[D^ra]=(-1)^r(\wt a)_r[a]$ make the series invariant under $z\mapsto-z/(1+z)$. The invariant coordinate is $w=z^2/(1+z)$, and the pole bounds at $-1,\infty$ give a polynomial in $w$ of degree at most $h$. Let $D_z=z\partial_z$, $D_w=w\partial_w$. Equation \eqref{H:Zhurec} specifies the scalar operator \begin{equation}\label{H:Lz} \mathcal L_z=\sum_{p=0}^9z^pK_p(D_z+p,h,x). \end{equation} Its symmetric version $\mathcal L_w$ is specified uniquely by \begin{equation}\label{H:coordinate} \mathcal L_z(P(w(z)))=(z+2)(z+1)^4(\mathcal L_wP)(w(z)). \end{equation} Thus $\mathcal L_wP=0$. Equations \eqref{H:Zhurec} and \eqref{H:coordinate} give an explicit finite definition of this order-nine operator. \subsection{The rank-three systems and the rational gauge} For $H=2t-1$ or $2/t-1$, consider \begin{equation}\label{H:rankthree} \left[D_w(D_w-H/2)(D_w-(3H+1)/2) +\frac w4(D_w-H)\bigl(D_w^2-(H-1)D_w-2(H+1)x\bigr)\right]F=0. \end{equation} For the first system, choose $\lambda^2=n+4x$ and $q=\sqrt t\lambda$. In the coordinate $-w/4$, its hypergeometric parameters are \begin{equation}\label{H:parameters} \alpha=(1-2t,1-t+q,1-t-q),\qquad \beta=(1,3/2-t,2-3t). \end{equation} The second system replaces $t$ by $t^{-1}$. Write $\mathcal H_t(x)$ and $\mathcal H_{1/t}(x)$ for their rank-three monodromy representations. At $w=-4$ the exceptional local exponent is always $1/2$, besides $0,1$. \begin{proposition}[Exact gauge identities]\label{H:gaugeclaim} The scalar row in Appendix~\ref{H:gauge} defines a rational gauge $\Gamma(w,x)$ from the tensor product of \eqref{H:rankthree} to $\mathcal L_w$, in derivative coordinates. Set \[ A=4x-3t+2,\quad B=4tx+2t-3,\quad C_t=4tx-3t^2+6t-3, \] and \[ \mathfrak d=(2t-1)^2(2/t-1)^2x(4x-3t+2)(4x-3/t+2)(8x-3h). \] Then $\Gamma_{\rm num}=\mathfrak d\Gamma$ is polynomial in $x$, and \begin{equation}\label{H:det} \det\Gamma=\frac{K(t)}{w^9(w+4)^9x^4A^2B^2C_t}, \end{equation} where \[ K(t)=\frac{5038848(t-2)^3(t-1)^{21}(t+1)^5(t+2)^2 (2t-3)^9(2t-1)^3(2t+1)^2(3t-2)^9}{t^{24}}. \] The residual ranks of $\Gamma_{\rm num}$ at $x=0,3n/4$ are four and one. For origin Frobenius solutions with exponents \[ \rho_0(t)=0,\qquad\rho_1(t)=t-1/2,\qquad\rho_2(t)=3t-1, \] the scalar row shifts the leading exponent of the product indexed by $(a,b)$ by exactly $-ab$. Its normalized leading coefficient for $(0,0)$ is one; for $(2,2)$ it is a nonzero function of $t$ divided by $\mathfrak d$, so has a simple pole at $x=3n/4$. \end{proposition} \begin{proof} The scalar row and the rank-three equations reduce the assertion to polynomial identities. Apply the scalar row to the nine tensors $(D_w^pF_1)(D_w^qF_2)$, $0\leq p,q\leq2$, and eliminate all higher derivatives with \eqref{H:rankthree}. Successive Euler derivatives through order eight give $\Gamma$. Substituting these rows into the operator specified by \eqref{H:Lz} and \eqref{H:coordinate}, and clearing denominators, gives zero in each of the nine tensor components. Thus the differential identity holds over $\Q(t,x,w)$. At the origin, the rank-three recurrences give the Frobenius coefficients. Substitution into the scalar row cancels every power preceding the stated shift and gives the nine nonzero leading coefficients, with the stated normalizations. The trace difference of the Euler first-order systems is $-18(w+2)/(w+4)$, giving the factor $w^{-9}(w+4)^{-9}$ in the determinant. The product of Frobenius leading coefficients and the ratio of the two Vandermonde determinants give its remaining factor. For the exceptional ranks, a nonzero $4\times4$ minor at $x=0,w=1$, using derivative rows $0,1,2,3$ and tensor columns $(1,1),(1,2),(2,1),(2,2)$, is \[ -\frac{820125(t-2)^6(t-1)^{11}(t+1)(2t-3)^8(2t-1)^6(3t-2)^8}{t^{20}}. \] At $x=3n/4,w=1$, the row-zero, column-$(0,0)$ entry is \[ \frac{441(t-2)^2(t-1)^4(2t-3)(2t-1)^2(3t-4)(3t-2)(4t-3)}{4t^6}. \] At zero the five columns with $p=0$ or $q=0$ vanish; at $3n/4$ the $2\times2$ minors vanish identically. The displayed nonzero certificates give the asserted ranks. All $t$-factors displayed above are nonzero, since $t>4$ at the parameters considered here. \end{proof} \subsection{The regular inverse and rational exceptional weights} We apply Proposition~\ref{H:gaugeclaim}. Over the discrete valuation ring in $x$ at zero, $\det\Gamma_{\rm num}$ has valuation five and residual rank four, so its elementary-divisor valuations are $0,0,0,0,1,1,1,1,1$. Since $\mathfrak d$ has a simple zero, $\Gamma^{-1}$ is regular and has residual rank five. At $x=3n/4$, the corresponding determinant valuation is eight and the residual rank is one; the inverse is again regular, now of rank eight. These statements are valid on the solution spaces at the origin as well. The scalar exponents are independent of $x$, with the only integral resonance class consisting of $0,n,3n$. For generic $x$ the gauge gives a log-free basis; the finitely many resonance obstructions are polynomial identities in $x$ with denominators independent of $x$, so vanish identically. Choosing the free resonant coefficients to be zero gives a basis holomorphic in $x$. In this basis \[ (\Gamma^{-1}P)_{00}=P(0) \] at both exceptional parameters and after base change to every local Artin algebra. At $x=3n/4$, the inverse kernel is the scalar Frobenius line beginning with $w^{3n}$: the $(2,2)$ coefficient has a simple pole there, and the inverse has a one-dimensional kernel. For rational $\xi$, either $\mathcal H_t(\xi)$ or $\mathcal H_{1/t}(\xi)$ can be reducible only at \begin{equation}\label{H:exceptional} \xi\in\mathscr E=\{0,1,3n/4,3n/4+4\}. \end{equation} To check this, apply the hypergeometric criterion $\alpha_i-\beta_j\notin\Z$ for irreducibility \cite[Propositions~3.2--3.3]{H:BH}. The differences involving $1-2t$ are irrational. An integral remaining difference forces $q=at+b$ with $a\in\{0,\pm1,\pm2\}$ and rational $b$. Using $q^2=t(n+4\xi)$ and $t^2=(n+2)t-1$, comparison in the basis $1,t$ gives $b^2=a^2$. The case $a=0$ comes from a nonzero half-integer $b$ and is impossible. The other cases give $n+4\xi\in\{n,m,4n,4m\}$, exactly \eqref{H:exceptional}. \subsection{Reduced local Zhu factors away from the exceptions} Let $\mathcal B_\xi$ be a local Artin factor of $A(V)$, with residue $x=\xi\in\Q\setminus\mathscr E$. Its polynomial \eqref{H:P}, under the inverse gauge, is a single-valued horizontal tensor, equivalently an intertwiner \[ T:\mathcal H_t(x)^*\longrightarrow\mathcal H_{1/t}(x) \] over $\mathcal B_\xi$. Monodromy matrices depend holomorphically on $x$ at a nonsingular base point, so substitution of their finite Taylor series makes this statement meaningful over the Artin algebra. The determinant of the gauge is a unit there. The reduction of $T$ is nonzero since $P(0)=\nu\one$; irreducibility and Schur's lemma make it an isomorphism. It remains an isomorphism over $\mathcal B_\xi$. Comparing the traces of monodromy at infinity gives the identity \begin{align} G(x)&=\cos(2\pi\sqrt{t(n+4x)}) -\cos(2\pi\sqrt{(n+4x)/t})\notag\\ &=-2\sin(\pi\sqrt m\sqrt{n+4x}) \sin(\pi\sqrt n\sqrt{n+4x})=0\quad\text{in }\mathcal B_\xi.\label{H:G} \end{align} This is entire in $x$, independent of the square-root choices. Its zeros are \[ \xi=\frac{a^2}{4n}-\frac n4\quad\text{or}\quad \xi=\frac{b^2}{4m}-\frac n4\qquad(a,b\in\Z). \] Every zero is simple as a zero in $x$. At $n+4x=0$, the product is a nonzero constant times $n+4x$; elsewhere both sine factors cannot vanish because $\sqrt{nm}$ is irrational. Thus $G(x)=(x-\xi)u(x)$ with $u(\xi)\ne0$, and \eqref{H:G} gives $x=\xi\one$ in the local algebra. With $x$ scalar, Schur's lemma makes the space of single-valued tensor solutions one-dimensional. The condition $P(0)=\nu\one$ fixes its multiple, so every coefficient $Z_d$ is scalar. Since $Z_h=Cy+b(x)$ with $C\ne0$, $y$ is scalar as well, and $\mathcal B_\xi=\C$. The same normalization determines the same $y$ at every factor over $\xi$; hence at most one maximal ideal lies over any such $\xi$. \subsection{The two upper exceptional parameters} At $x=1$, the numerator parameters are $(1-2t,-2t,2)$. Hypergeometric contiguity \cite[Propositions~2.5 and~2.7]{H:BH} gives \[ 0\longrightarrow R_t\longrightarrow\mathcal H_t(1) \longrightarrow\one\longrightarrow0, \] with $R_t$ irreducible of rank two. There is no trivial subrepresentation: its scalar solution would be a rational function regular at $w=0,-4$ and with only integral infinity exponent $-2$, hence a polynomial of negative degree. Consequently any map from $\mathcal H_t(1)^*$ to $\mathcal H_{1/t}(1)$ kills the trivial source submodule and has image in $R_{1/t}$. Its $(0,0)$ origin coefficient is zero. The only integral-exponent products are $(0,0),(1,1),(2,2)$, and the other two become series starting at $w^n,w^{3n}$ under the gauge. Thus $P(0)=0$, contradicting \eqref{H:P}. At $x=3n/4+4$, the numerator parameters are $(1-2t,t+3,-3t-1)$, giving \[ 0\longrightarrow\chi_t\longrightarrow\mathcal H_t(x) \longrightarrow R_t\longrightarrow0. \] Here $\chi_t$ has origin exponent $3t-1$. This sequence is nonsplit: a quotient $\chi_t$ would give an adjoint solution with character $\chi_t^{-1}$, equal to $w^{-3t+1}$ times a polynomial of degree $-3$. Indeed its infinity exponent is $-3t-2$, and its exponents at $-4$ are $0,1,1/2$. An intertwiner from the dual of the first system to the second must therefore factor through the rank-one source quotient into the rank-one target submodule. Its only origin component is $(2,2)$, which the scalar row sends to a series starting at $w^{3n}$. It cannot give a nonzero polynomial of degree at most $h=2n$. Neither upper exceptional value is a Zhu eigenvalue. \subsection{The two lower exceptional parameters} Let $\mathcal B_\xi$ now be a local factor at $\xi=0$ or $3n/4$. The regular inverse gauge produces an intertwiner $T$ whose reduction is nonzero, because its $(0,0)$ coefficient is $P(0)=\nu\one$. At zero the numerator parameters are $(1-2t,2-2t,0)$, giving the nonsplit sequence \[ 0\longrightarrow\one\longrightarrow\mathcal H_t(0) \longrightarrow R_t\longrightarrow0. \] A trivial quotient would yield a rational adjoint solution regular at the two finite singularities and with infinity exponent $-1$, which is impossible. A nonzero intertwiner from the dual of the first system to the second has rank one and factors through their trivial constituents. At infinity the eigenvalue one is isolated from the other two eigenvalues, both $e^{-4\pi it}$ in the first representation. Its spectral projector varies holomorphically in $x$. The component of $T$ between the matching one-dimensional eigenlines is a unit over the Artin algebra. Equality of the eigenvalues therefore forces \[ e^{2\pi i s(x)}-1=0,\qquad s(x)=-n+\sqrt n\sqrt{n+4x}. \] Since $s(0)=0$ and $s'(0)=2$, this gives $x=0$ in $\mathcal B_0$. At $x=3n/4$, the parameters are $(1-2t,t-1,3-3t)$, and \[ 0\longrightarrow R_t\longrightarrow\mathcal H_t(3n/4) \longrightarrow\chi_t\longrightarrow0 \] is nonsplit. A $\chi_t$ submodule would give $w^{3t-1}$ times a polynomial with infinity exponent $3t-3$, hence degree $-2$. A nonzero intertwiner between the dual and the other system factors as an isomorphism of the irreducible rank-two constituents. All infinity eigenvalues are distinct, and one of the matched moving eigenline components is a unit. The exponent difference is again $s(x)$; now $s(3n/4)=n$ and $s'(3n/4)=1$. Thus $x=(3n/4)\one$ in its local Artin factor. To finish, at $x=0$ the Zhu quadratic is $y(y-w_0)=0$ with $w_0\ne0$, and both roots are realized by $V$ and $X$. The whole summand is therefore $\C\oplus\C$. At $x=3n/4$, the inverse-gauge kernel begins at $w^{3n}$ and meets no polynomial of degree at most $2n$. Such polynomials inject into the one-dimensional homomorphism space of the rank-two constituents. Normalization fixes the polynomial uniquely, and $Z_h=Cy+b(x)$ makes $y$ scalar. The current $J$ realizes this factor, which is exactly $\C$. \begin{theorem}[Uniform Zhu semisimplicity]\label{H:Zhuconclusion} For every odd $n\geq3$, the Zhu algebra $A(V_{n+2})$ is semisimple. The zero-weight summand is $\C\oplus\C$, each nonzero weight fibre has dimension at most one, and weight one is absent. Consequently $V_{n+2}$ is strongly rational. \end{theorem} \begin{proof} Theorem~\ref{H:C2bound} gives uniform $C_2$-cofiniteness. Rationality of conformal weights under $C_2$-cofiniteness \cite[Corollary~5.10]{H:Miyamoto} makes every residue of $x$ rational. At a rational value outside \eqref{H:exceptional}, the other factors in \eqref{H:det} do not vanish: $A,B$ could vanish only at irrational $x$, and $C_t$ vanishes at $3n/4$. The cases just treated therefore exhaust all local Artin factors, proving semisimplicity. The simple, self-contragredient CFT-type VOA then satisfies McRae's rationality criterion \cite[Main Theorem~3]{A:McR}. This proves strong rationality. \end{proof} The argument uses the uniform identities above and is independent of the character classification and categorical reconstruction in the main text. \section{Complete coefficients of the uniform Zhu gauge}\label{H:gauge} This appendix gives all thirty-nine nonzero scalar-row coefficients of the rational gauge. Together with the thirty singular-vector coefficients in Appendix~\ref{app:null} and the fixed recursions in Appendix~\ref{H:uniform}, it specifies every rational function in Proposition~\ref{H:gaugeclaim}. The symbols $E_j$ in this appendix denote auxiliary polynomials. For solutions $F_1,F_2$ of the rank-three equations \eqref{H:rankthree}, the scalar row is \begin{equation}\label{H:scalarrow} \frac1{\mathfrak d}\sum_{r=-4}^{0}\sum_{p,q=0}^{2} R_{r,p,q}(t,x)w^r(D_w^pF_1)(D_w^qF_2), \end{equation} where $\mathfrak d$ is as in Proposition~\ref{H:gaugeclaim}. Put \[ \begin{gathered} \ell_1=t-2,\quad\ell_2=t-1,\quad\ell_3=2t-3,\quad\ell_4=2t-1,\\ \ell_5=3t-4,\quad\ell_6=3t-2,\quad\ell_7=4t-3. \end{gathered} \] All omitted coefficients vanish. The table is grouped by the Laurent power $r$. {\small\renewcommand{\arraystretch}{1.7} \begin{longtable}{@{}rrr l@{}} \toprule $r$&$p$&$q$&$R_{r,p,q}(t,x)$\\\midrule \endfirsthead \toprule $r$&$p$&$q$&$R_{r,p,q}(t,x)$\\\midrule \endhead $-4$&1&1&$-36\ell_1^2\ell_3^2\ell_4^2\ell_5\ell_6^2\ell_7/t^5$\\ $-4$&1&2&$-72\ell_1\ell_3^2\ell_4^2\ell_5\ell_6^2\ell_7/t^4$\\ $-4$&2&1&$72\ell_1^2\ell_3^2\ell_4\ell_5\ell_6^2\ell_7/t^5$\\ $-4$&2&2&$144\ell_1\ell_3^2\ell_4\ell_5\ell_6^2\ell_7/t^4$\\\addlinespace $-3$&0&1&$-72\ell_1^2\ell_3^2\ell_4^2\ell_5\ell_6\ell_7x/t^4$\\ $-3$&0&2&$-144\ell_1\ell_3^2\ell_4^2\ell_5\ell_6\ell_7x/t^3$\\ $-3$&1&0&$72\ell_1^2\ell_3\ell_4^2\ell_5\ell_6^2\ell_7x/t^5$\\ $-3$&1&1&$36E_1\ell_1\ell_2^2\ell_3\ell_4\ell_6/t^5$\\ $-3$&1&2&$-18E_2\ell_1\ell_3\ell_4\ell_6/t^4$\\ $-3$&2&0&$-144\ell_1^2\ell_3\ell_4\ell_5\ell_6^2\ell_7x/t^5$\\ $-3$&2&1&$-18E_3\ell_1\ell_3\ell_4\ell_6/t^5$\\ $-3$&2&2&$72E_4\ell_1\ell_3\ell_4\ell_6/t^4$\\\addlinespace $-2$&0&0&$144\ell_1^2\ell_3\ell_4^2\ell_5\ell_6\ell_7x^2/t^4$\\ $-2$&0&1&$12E_5\ell_1\ell_3\ell_4^2\ell_6x/t^4$\\ $-2$&0&2&$-12E_6\ell_1\ell_3\ell_4^2\ell_6x/t^3$\\ $-2$&1&0&$12E_7\ell_1^2\ell_3\ell_4\ell_6x/t^5$\\ $-2$&1&1&$6E_8\ell_1\ell_3\ell_4\ell_6/t^5$\\ $-2$&1&2&$3E_9\ell_1\ell_3\ell_4\ell_6/t^4$\\ $-2$&2&0&$-12E_{10}\ell_1^2\ell_3\ell_4\ell_6x/t^5$\\ $-2$&2&1&$-3E_{11}\ell_1\ell_3\ell_4\ell_6/t^5$\\ $-2$&2&2&$3E_{12}\ell_1\ell_3\ell_4\ell_6/t^4$\\\addlinespace $-1$&0&0&$48E_{13}\ell_1^2\ell_3\ell_4^2\ell_6x^2/t^4$\\ $-1$&0&1&$6E_{14}\ell_1\ell_3\ell_4^2\ell_6x/t^4$\\ $-1$&0&2&$-6E_{15}\ell_1\ell_3\ell_4^2\ell_6x/t^3$\\ $-1$&1&0&$-6E_{16}\ell_1^2\ell_3\ell_4\ell_6x/t^5$\\ $-1$&1&1&$-6E_{17}\ell_1\ell_3\ell_4\ell_6/t^5$\\ $-1$&1&2&$3E_{18}\ell_1\ell_3\ell_4\ell_6/(2t^4)$\\ $-1$&2&0&$-6E_{19}\ell_1^2\ell_3\ell_4\ell_6x/t^5$\\ $-1$&2&1&$-3E_{20}\ell_1\ell_3\ell_4\ell_6/(2t^5)$\\ $-1$&2&2&$3E_{21}\ell_1\ell_3\ell_4\ell_6/(2t^4)$\\\addlinespace 0&0&0&$6E_{22}\ell_1^2\ell_3\ell_4^2\ell_6x/t^4$\\ 0&0&1&$3E_{23}\ell_1\ell_3\ell_4^2\ell_6x/t^3$\\ 0&0&2&$-3\ell_1\ell_2\ell_3\ell_4^2\ell_6x(16x-1)/t^2$\\ 0&1&0&$-3E_{24}\ell_1^2\ell_3\ell_4\ell_6x/t^4$\\ 0&1&1&$-3E_{25}\ell_1\ell_3\ell_4\ell_6x/t^4$\\ 0&1&2&$3E_{26}\ell_1\ell_3\ell_4\ell_6x/t^3$\\ 0&2&0&$-3\ell_1^2\ell_2\ell_3\ell_4\ell_6x(16x-1)/t^4$\\ 0&2&1&$-3E_{27}\ell_1\ell_3\ell_4\ell_6x/t^4$\\ 0&2&2&$3E_{13}\ell_1\ell_3\ell_4\ell_6x/t^3$\\ \bottomrule \end{longtable}} The twenty-seven auxiliary polynomials are as follows. Each display is an exact expression over $\Z[t,x]$. \begin{align*} E_1={}&80t(t^2-t+1)x-13(t-2)(2t-3)(2t-1)(3t-2),\\ E_2={}&16t(t-1)(4t^2+2t-7)x\\ &+(2t-3)(2t-1)(3t-2)(40t^2-87t+46),\\ E_3={}&16t(t-1)(7t^2-2t-4)x\\ &-(t-2)(2t-3)(3t-2)(46t^2-87t+40),\\ E_4={}&40t(t-1)^2x+(2t-3)(3t-2)(17t^2-35t+17),\\ E_5={}&4t^2(16t-17)x-(t-2)(2t-3)(54t^2-97t+42),\\ E_6={}&8t(8t-9)x+(2t-3)(72t^2-143t+66),\\ E_7={}&4t(17t-16)x+(2t-1)(3t-2)(42t^2-97t+54). \end{align*} \begin{align*} E_8={}&8t^2(8t^2-11t+8)x^2\\ &+2t(108t^4-308t^3+395t^2-308t+108)x\\ &-27(t-2)(t-1)^2(2t-3)(2t-1)(3t-2),\\ E_9={}&32t^2(4t-7)x^2-8t(60t^3-110t^2+29t+18)x\\ &-3(t-1)(2t-3)(2t-1)(3t-2)(22t-27),\\ E_{10}={}&8t(9t-8)x+(3t-2)(66t^2-143t+72),\\ E_{11}={}&32t^2(7t-4)x^2+8t(18t^3+29t^2-110t+60)x\\ &-3(t-2)(t-1)(2t-3)(3t-2)(27t-22),\\ E_{12}={}&64t^2x^2+16t(39t^2-79t+39)x\\ &+3(2t-3)(3t-2)(32t^2-65t+32),\\ E_{13}={}&4tx+9t^2-19t+9,\\ E_{14}={}&32t^2x^2+8t(t^2+5t-7)x\\ &-(t-2)(t-1)(36t^2-79t+36). \end{align*} \begin{align*} E_{15}={}&4t(16t-17)x+36t^3-124t^2+134t-45,\\ E_{16}={}&32t^2x^2-8t(7t^2-5t-1)x\\ &-(t-1)(2t-1)(36t^2-79t+36),\\ E_{17}={}&32t^3x^3-8t^2(11t^2-19t+11)x^2\\ &+t(12t^4-95t^3+170t^2-95t+12)x\\ &+3(t-2)(t-1)^2(2t-3)(2t-1)(3t-2),\\ E_{18}={}&16t^2(8t-13)x^2-4t(84t^3-196t^2+128t-21)x\\ &-3(t-1)(2t-3)(2t-1)(3t-2)(4t-5),\\ E_{19}={}&4t(17t-16)x+45t^3-134t^2+124t-36,\\ E_{20}={}&16t^2(13t-8)x^2-4t(21t^3-128t^2+196t-84)x\\ &-3(t-2)(t-1)(2t-3)(3t-2)(5t-4),\\ E_{21}={}&64t^2x^2+8t(33t^2-68t+33)x +15(t-1)^2(2t-3)(3t-2). \end{align*} \begin{align*} E_{22}={}&8tx^2+2(t^2-3t+1)x+(t-1)^2,\\ E_{23}={}&16tx^2-4(3t^2-9t+7)x+(t-1)(3t-4),\\ E_{24}={}&16tx^2-4(7t^2-9t+3)x+(t-1)(4t-3),\\ E_{25}={}&16t^2x^2-4t(9t^2-16t+9)x\\ &+18t^4-78t^3+121t^2-78t+18,\\ E_{26}={}&4t(2t-3)x-18t^3+46t^2-36t+9,\\ E_{27}={}&4t(3t-2)x-9t^3+36t^2-46t+18. \end{align*} For completeness, the change of differential coordinates in \eqref{H:coordinate} can be recovered without solving for an unspecified operator. 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